Log Equations

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A. I. Prilepko Solution

Logarithmic equation Page number 92 (Q. No. 125 to 261) 125.

⇒ ⇒ ⇒

x – 1 ≠ 1 and x≠2

logx – 1 3 = 2 (x – 1)2 = 3

x–1>0 x>1

x2 + 1 – 2x = 3 x2 – 2x – 2 = 0 D = b2 – 4ac = 12 Roots =

−b+ D 2±2 3 2(1 ± 3 ) = = =1± 2a 2 2

3 ∴ Final solution x ∈ {

Since 1+ 3 satisfies the equation

}

126.

log4(2 log3(1 + log2(1 + 3 log3x))) =

1 2

⇒

1 1 log2(2 log3 (1 + log2 (1 + 3 log3x))) = 2 2

⇒ ⇒ ⇒ ⇒

2 log3 (1 + log2(1 + 3 log3x))) = 21 = 2

⇒ ⇒ ⇒

(1 + log2 (1 + 3 log34) = 31

127.

log3 (1 + log3 (2x – 7) = 1

log2 (1 + 3 log3x) = 3 – 1 3 log3x = 3 x=

31

1 + 3 log3x = 22 log3x = 1 ∴ Final solution x ∈ {3}

=3

⇒

2x = 9 + 7 = 16

128.

log3 (3x – 8) = 2 – x

⇒

⇒ ⇒

32 – x = 3x – 8

⇒

9 = t2 – 8t

(2x – 7) = 32

∴ Final solution x

x=4

Let 3x = t 9 =t–8 t

⇒

1 + log3 (2x – 7) = 3

⇒

{4}

⇒

= 3x – 8

9 3x

= 3x – 8

∈12 log 3 2 (39 − 2 x ) 1+ 3 2t2x –38t − x– 9 = 0 3

⇒

D = b2 – 4ac = 64 + 36 = 100 t = 3x =

8 ± 10 18 2 = ,– = 9 ,–1 (negative value is not possible as ax >0 ) 2 2 2

3x = 9 ⇒ 3x = 32 ⇒ x = 2 129.

x≠3

=1

⇒ log2 (9 – 2x) = 3 – x Now,

2x

∴ Final solution x ∈ {2}

= 1, 8

⇒(t – 1) (t – 8) = 0

23

⇒ 9 – 2x = (2)3 – x

⇒ 9 – 2x =

⇒ 8 = 9t – ⇒t = 1, 8

⇒ – 9t + 8 = 0 ⇒ x = 0, 3 but x ≠ 3

t2

2

⇒

x

8 =9–t t

t2

∴ Final solution x ∈ {0} 130.

⇒ 131.

25 – 65 = 8x

⇒ (5 – x)2 = x2 – 2x + 65 ⇒ – 40 = 8x

log3 (log9x +

+ 9x) = 2x

∴ Final solution x {– 5} 1 1 1 ⇒ log3 (log9x + + 32x) = 2x ⇒ – = log3x 2 2 2

–1 = log3x

⇒x=

log5 – x (x2 – 2x + 65) = 2

1 3

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⇒ 25 – x2 – 10x = x2 – 2x + 65 ⇒x=–5

∴ Final solution x ∈ {

}

A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics 132.

⇒ ⇒

log3(x + 1) (x + 3) = 1

A. I. Prilepko Solution

31 = (x + 1) (x + 3)

3 = x2 + 4x + 3 0 = x2 + 4x ∴ x = –4, 0 since – 4 doesn’t satisfy equation 133.

134.

log7 (2x – 1) + log7 (2x – 7) = 1 Let 2x = t log7(t – 1) (t – 7) = 1 ⇒ 71 = t2 – 7t – t + 7

⇒ t = 0, 8

2x = 8 ⇒ x = 3 2x = 0 Impossible

∴ Final solution x

log 5 + log(x + 10) – 1 = log (2/x – 20) – log (2x – 1)

⇒ log (5 + x + 10) – log1010 = log

5( x + 10 ) 21x − 20 = log 10 2x − 1 ⇒ (2x – 1) (x + 10) = 42x – 40 ⇒ 2x2 – 19x – 42x – 10 + 40 = 0

∴ Final solution x ∈ {

x + 10 21x − 20 = 2 2x − 1 ⇒ 2x2 – x + 20x – 10 = 42x – 40 ⇒ 2x2 – 20x – 3x + 30 = 0 3 ⇒ x = , 10 2

, 10 }

1 1  1   log + log x + log 5  2 3 3  

135.

1 – log 5 =

⇒

log10

⇒

1 log1023 = log10  x 2

⇒ log10 10 – log105 =

10 1 = [log102–1 + log10x + log1051/3] 5 3 3

x−

x35 ∈ 1⇒x8− =20 321 16 2 3 23 25x − 1

⇒x=

= log

1 2 1 x+ 8

  ⇒    

  1  x− 2  

2

x

1 x2 x x x2 = x3 + + x2 – – – 8 8 4 2 2

⇒

x3 +

⇒

8x3 + x2 = 8x3 – 2x + 4x2 – 1

⇒

0 = 3x (x –1) +1(x – 1)

137.

3log3 log

⇒

log10 x – log10x + log102x – 3 = 0

⇒

⇒

log10x – 2 log10x + 2(log10x) 2 – 62 = 0

⇒

⇒

2(log10x)2 – 4 log10x + 3 log10x – 6 = 0 ⇒

⇒

log10x = 2

or

log10x = –

⇒

x = 102 = 100

or

x = 10–3/2

x

16 3

5

}

x+

1 2

1 [log102–1 + log10x + log1051/3] 3

1 log102–1 + log10x + log1051/3 3

1 1 1    1 1 log  x −  = log  x +  – log  x +  ⇒log x – log 2 2 8    2 2

x

log

⇒ log1023 =

 5 

∴ Final solution x ∈ {

log x –

{3}

⇒

⇒ (2x – 3) (x – 10) = 0

⇒

∴ Final solution x ∈ {0}

⇒ 0 = t2 – 8t

⇒ log

136.

⇒ 3 = x2 + x + 3x + 3 ⇒ 0 = x(x + 4)

=

   x+ 1   2    1  x +  8  

1 1 = log x + – log 2 2

2

⇒

1 +x 4 1 x+ 8

x2 +

x2 x−

x+

1 2

=

⇒ 8(8 x3 + x2) = 8(8x3 + 2x + 8x2 – 4x2 – 1 – 4x) ⇒ 0 = 3x2 – 2x – 1 1 ⇒x=– ,1 3 ⇒

– logx + log2x – 3 = 0

x−

⇒ 0 = 3x2 – 3x + x – 1 ∴ Final solution x ∈ { –

}

log x – log x + log2x – 3 = 0 1 log10x – log10x + (log10x)2 – 3 = 0 2 2(log10x)2 – log10x – 6 = 0

(2 log10x + 3) (log10x – 2) = 0 3 2

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A. I. Prilepko Solution

1 8

Mathematics Online: It’s all about Mathematics 2

( x − 2 ) + log( x − 2 ) 5 −12

A. I. Prilepko Solution

138.

( x − 2) log

⇒

log ( x − 2)log

⇒ ⇒ ⇒ ⇒

log (x – 2) (log2(x – 2)+ log (x – 2)5 – 12)) = log10102 log(x – 2) t [t2 + 5t – 12)] = 2t ⇒ t3 + 5t2 – 14t = 0 ⇒ t(t + 7)(t – 2) = 0 ⇒ t = 0, –7, 2 ⇒ x = 2 ,2 + 10–7 , 102

139.

9 log3 (1− 2 x ) = 5x2 – 5

⇒

3 2 log3 (1− 2 x ) = 5x2 – 5

⇒ ⇒ ⇒ ⇒ ⇒

3log3 (1−2 x ) = 5x2 – 5

2

(Let log (x-2) = t )

= 102 log (x – 2)

( x − 2 ) + log( x − 2 )( 5 −12 )

= log 102log (x – 2) t(t2 + 5t – 14) = 0 log (x-2) = 0, –7, 2

2

(1 – 2x)2 = 5x2 – 5 1 + 4x2 – 4x = 5x2 – 5 0 = 5x2 – 4x2 – 4x – 5 – 1 0 = x2 – 4x – 6 D = b2 – 4ac = 16 + 24 = 40 D = 2 10 α=

140.

⇒ ⇒ ⇒

⇒ ⇒ ⇒ ⇒

10

x1 + log x = 10x logxx1 + logx = log 10x (log10x) (1 + log10x) = log1010 + log10x ⇒ log10x + log102x = 1 + log10x log102x = 1 ∴

141.

−b+ D = 2 + 10 , 2 – 2a

log10 x = ±1 x = 10–1, 101

x2 logx = 10x2 log10x2.log x = log1010 + log10x2 2 log10x (log10x) = 1 + 2 log10x 2 log102x – 2 log10x – 1 = 0

Let

log10x = t

2t2 – 2t – 1 = 0 2t2 D = b2 – 4ac = 4 + 8 = 12 t=

D = 2 3

2(1 ± 3 ) 1± 3 2±2 3 = = 4 2 9

Now, log10x = t =

1+ 3 1− 3 , 2 2

1± 3 2

∴

x=

⇒

log10x =

⇒

log10x2 = 1± 3

1± 3 2

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A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics

⇒

x2 = 101±

142.

x

log x +5 3

log x .

3

,

101−

log x + 5 = log10105 + log10x 3 log x + 5 = 5 + log10x 3

⇒

log x .

⇒ ⇒ ⇒

log2x + 5 log10x = 15 + 3 log10x log102x + 2 log10x – 15 = 0 log10x (log10x + 5) – 3(log10x + 5) = 0 (log10x – 3) (log10x + 5) = 0 log10x = 3 x=

log10x = – 5

103

x = 10–5

143.

x log3 . x = 9

⇒ ⇒ ⇒

log3 x log 3.x = log39 log3x (log3x) = 21

⇒

log3x = ±

⇒

x= 3

144.

( x )log5 x −1 = 5

⇒

log5

⇒

1 log5x (log5x – 1) = 1 2

⇒

1 1 log52x – log5x = 1 2 2

⇒ ⇒ ⇒ ⇒ ⇒

= log52 x – log5x = 2

log5x = 2

and

⇒

x = 52

and

⇒

x=5

and

145.

xlogx +1 = 106 log x (log x + 1) = log10106

⇒ ⇒ ⇒ ⇒

3

= 105 + log x

⇒ ⇒

⇒

3

101+

x=

A. I. Prilepko Solution

log32x = 21

2

2

, 3−

2

x . log5 (x – 1) = log55

= log52 x – log5x – 2 = 0 log5x (log5x + 1) – 2(log5x + 1) = 0 (log5x – 2) (log5x + 1) = 0 log5x = –1 1 x= 5 1 x= 5

log102x + log10x – 6 = 0 (log10x – 2) (log10x + 3) = 0 log10x = 2

log10x = – 3

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A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics ⇒ x = 102 x = 10–3

146.

log x +7 x 4

⇒

log x. log

⇒

2 log10 x + 3 log10x – 4 = 0

⇒

2 log10 x + 4 log10x – log10x – 4 = 0

⇒ ⇒

(log10x + 4) (log10x – 1) = 0

147.

x log

⇒

x1/ 2 logx ( x −2 ) = 9

⇒

x logx ( x −2 ) = 9

⇒

( x − 2) 2 log x x = 9

= 10log x + 1 x+7 = 4 log10 x + 4 4

x = 104 and x

A. I. Prilepko Solution

( x −2)

x = 101

=9

2

x

⇒⇒ (x – 2)2 – 9 = 0 ⇒x2 + 4 – 4x – 9 = 0 ⇒ x2 – 4x – 5 = 0 ⇒ x(x – 5) + 1(x – 5) = 0 ⇒ x = –1, 5 but x > 0 ⇒ ∴ x= 5 is a solution log2 x + log x2 − 2

148.

 log x     2 

⇒

2 2 1 1 log x log x + log x − 2 = 2 log x 2 2

⇒

 log x     2 

= log x

log2 x + log x 2 − 2

= log

 log x   ⇒  2 

149.

x

log2 x + log x 2 − 2

= 2 log x

3 log 2 x – log28x + 1 = 0

⇒ 3 log 2 x – (log223 + log2x) + 1 = 0 ⇒ 3 log 2 x – 3 – log2x + 1 = 0 ⇒ 3 log2 x – 2 – log2x = 0 2

  ⇒  3 log2 x  = (2 + log2x)2

⇒9 log2x = 4 + 4 log2x + log 22 x

⇒

⇒

9 log2x – 4 log2x = 4 + log22x 0 = log22x – 5 log2x + 4

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A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics ⇒ 0 = log22x – log2x – 4 log2x + 4

⇒

0 = log2x (log2x – 1) – 4(log2x – 1) log2x = 4 = 16 =2 log2x = 1

∴

150.

⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒

A. I. Prilepko Solution

log2x – 3 log x = log (x2) = 4 log2x – 3 log x – 2 log x + 4 = 0 log102x – 5 log10x + 4 = 0 log102x – log10x – 4 log10x + 4 = 0 (log10x – 4) (log10x – 1) = 0 log10x = 4 x=

log10x = 1

104

x = 102

151.

log1/3x – 3 log1 / 3 x + 2 = 0

⇒

– log3x – 3 − log 3 x + 2 = 0

⇒

  (2 – log3x)2 =  3 − log3 x   

⇒ ⇒ ⇒ ⇒ ⇒

4 + log32x – 4 log3x = – 9 log3x

⇒

x=

152.

2 (log x 5 )2 – 3 logx 5 +1 = 0

2

log32 x + 5 log3x + 4 = 0 (log3x + 4) (log3x + 1) = 0 log3x = – 4

log3x = – 1

3–4/3 =

3–1 = x

x

1 1 , 81 3

2

⇒

(log 2

⇒

x)

5

2 log52

2

–

3 +1=0 log 5 x

2 – 2 log x + 1 = 0 x 5

⇒ ⇒

0 = 2 log52x – 3 log5x + 1

⇒

log5x =

⇒

x=

153.

log22 x + 2 log2x – 2 = 0

⇒

log22 x + log2x – 2 = 0

⇒ ⇒ ⇒

(log2x – 1) (log2x + 2) = 0

0 = 2 log5x(log5x – 1) – 1(log5x – 1) 1 2

5

log2x = 1 x=

21

log5x = 1 x=5

log2x = – 2 x = 2–2

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A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics 1 4

⇒

=2

154.

(a logb x )2 – 5 x logb x + 6 = 0

=

A. I. Prilepko Solution

Let x logba = t

⇒

( x logb a )2 – 5 x logb a + 6 = 0

⇒ ⇒⇒ ⇒

t2 – 5t + 6 = 0 t2 – 2t – 3t + 6 = 0 t(t – 2) – 3(t – 2) = 0 (t – 3) (t – 2) = 0

⇒

log a t=3= x b

⇒

t=2= log 2 = x = 2logb a

x= 1 x

155.

log2 (100x) + log2 (10x) ± 14 + log

⇒

2 2 2 2 log10 10 2 + log10 x + log10 10 + log10 x = 14 + log

⇒

4 + log102 x + 1 + log102 x = 14 + log

156.

log4 (x + 3) – log4 (x – 1)

⇒

log4

x+3 = 2 – log 22 23 x −1

⇒

log4

x+3 3 =2– x −1 2

⇒

log4

x+3 1 = x −1 2

⇒

41/2 =

⇒ ⇒ ⇒

2(x – 1) = (x + 3)

157.

log4(x2 – 1) – log4(x – 1)2 = log4 4 − x 2

1 x

1 x logbxa 3 xlog logb a

x+3 x −1

2x – 2 = x + 3 x=5

1 ( x − 1)( x + 1) (4−x2 ) 2 = log 4 ( x − 1)( x − 1)

⇒

log4

⇒

log4

⇒

2  x + 1   =  4 − x 2     x − 1

x +1 = log4 (4 – x2)1/2 x −1 2

Mathematics Online: It’s all about Mathematics

A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics x 2 + 1 + 2x x 2 + 1 − 2x

158.

2 log3

= 4−x

A. I. Prilepko Solution

2

x−3 x −3 + 1 = log3 x−7 x −1 2

x−3  x−3   + 1 = log3  log3  x − 7  x −1   2

x−3  x −3  =–1  – log3  log3   x −1  x −7

log3

( x − 3) 2 ( x − 7)

2

( x − 3)( x − 1) ( x − 7) 2

×

x −1 =–1 x −3

= 3–1

x 2 − 4x + 3 2

x − 14 x + 49

=

1 3

3x2 – 12x + 9 = x2 + 14x – 49 ⇒ 2x2 + 2x – 40 = 0 ⇒ x2 + x – 20 = 0 ⇒ x (x + 5) – 4(x + 5) = 0 (x – 4) (x + 5) = 0 x = 4, – 5 since 4 can’t satisfied the equation ∴ – 5 is a solution 159.

2 log4 (4 – x) = 4 – log2(–2 – x) 2 log22 (4 – x) + log2(–2 – x) = 4 log2 (4 – x)(–2 – x) = 4 24 = x2 – 2x – 8 0 = x2 – 2x – 8 – 16 0 = (x – 6) (x + 4) ∴ x = 6, –4

160.

3 + 2 logx + 1 3 = 2 log3(x + 1) Let log3(x + 1) = t

⇒

2 3 + log ( x + 1) = 2 log3(x + 1) 3

⇒

=3+

⇒ ⇒ ⇒ ⇒ ⇒

3t + 2 = 2t2

2 = 2t t

0 = 2t2 – 3t – 2 0 = 2t2 – 4t + t – 2 0 = 2t (t – 2) + l (t – 2) 0 = (2t + 1) (t – 2) ∴

log3(x + 1) = − x + 1 = 3–1/2

1 2

or log3x + 1 = 2 x + 1 = 32

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A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics x=

1

–1

3

A. I. Prilepko Solution

x=8

 3 +3   2–  3    161.

logx 9x2 . log32x = 4 2 (logx9 + logxx2) log3 x = 4

(logx32 + 2) log32x = 4 (2 logx 3 + 2) (log32x) = 4 2 2 log3 x × log32x + 2 log3 x – 4 = 0

2 log3x + 2 log32 x – 4 = 0 2 log32 x + 4 log3x – 2 log3x – 4 =0 (2 log3x – 2) (log3x + 2) = 0 log3x = –2 or log3x = 1 1 9

x=3

log12/ 2

 x2  (4x) + log2  8 

x=

162.

   =8 

– log22 4x + log2x2 – log28 = 8 – (log22 4 + log22x) + 2 log2x – 3 = 8 – 2 log222 – log22x + 2 log2x – 11 = 0 163.

log 0.5 x x2 – 14 log16xx3 + 40 log4x

x =0

10

164.

6 – (1 + 4.9

4 − 2 log

4  6 – 1 + 4.9 4 log  9 39 

3

3

) . log7 x = logx7

1   log7x = log x  7 

 4.36 6 – 1 + 2 log3 3 4  3

1   log x = 7  log7 x 

36   6 – 1 + 4. 8  3  

165.

log3 (4.3x – 1) = 2x + 1 32x + 1 = 4.3x – 1 32x . 31 = 4.3x – 1 Let 3x = t 1 3

= 3t2 – 4t – 1 = 0

3x =

3t2 – 3t – t + 1 = 0 (3t – 1) (t – 1) = 0

3x = 1

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A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics t= 166.

1 ,1 3

A. I. Prilepko Solution

x=0

log3 (3x – 6) = x – 1 3x – 1 = 3x – 6 3x = 3x – 6 3

3x = 9 x = 22

t =t–6 3

t = 3t – 18 – 2t = – 18 t=9 167.

168.

log3 (4x – 3) + log3 (4x – 1) = 1 Let 4x = t log3(4x – 3) (4x – 1) = 1 31 = (t – 3) (t – 1) 3 = t2 – 4t + 3 0 = t (t – 4) t = 0, 4

Not possible 4x = 4 x=1

log3 (log1/22x – 3 log1/2x + 5) = 2 log 22 −1 x – 3 log2 −1 x + 5 = 32

⇒

∴

– log22x + 3 log2x = 4 0 = log22x + 3 log2x + 4 0 = (log2x – 1) (log2x + 4) log2x = 4 log2x = 4 x=2 x = 16 2

169.

log5

2+ x = log 5x +1 10

log5

2+x 2 – log5 =0 10 x +1

2 x +1 × x +1 2 log5

=0

2x + x 2 + 2 + x = 50 20

x2 + 3x + 2 = 20 x2 + 3x – 18 = 0 + 6x – 3x (x + 6) (x – 3) = 0 x = 3, – 6 Not satisfied 3 Ans. 170.

1 + 2 logx + 25 = log5 x + 2 Let log5x + 2 = t

5–1 = x + 2 x=

1 9 –2=– 5 5

Mathematics Online: It’s all about Mathematics

A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics 2 1 + log x + 2 = log5 x + 2 5

1+

2 =t t

172.

log5x + 2 = 2 52 = x + 2

t + 2 = t2 0 = t2 – t – 2 0 = (t – 2)–2 +1 (t + 1) t = –1, 2 171.

x = 23

log424x = 2log2 4 log424x = 4 24x = 44 44x = (22)4 24x = 8 4x = 8 x=2

log2

15 = log x − 1 2 8

Let log2x = t

⇒

15 log2x – log24 = log x − log 8 − 1 2 2

⇒

15 log2x – 2 = log x − 3 − 1 2 x   4

15 log2x – 2 = log x − 4 2

15 ( t − 4)

(t – 2) =

(t – 2) (t – 4) = 15 t2 – 2t – 4t + 8 = 15 t2 – 6t + 8 – 15 = 0 t2 – 6t – 7 = 0 t2 – 7t + t – 7 = 0 t (t – 7) + l(t – 7) = 0 (t + 1) (t – 7) = 0 t = –1, 7 log2x = –1 or x=

173.

A. I. Prilepko Solution

1 2

1 − 2(log x 2 )2 log x − 2(log x )2

log2x = 7 x = 27

=1

2 1 − 2( 4 log10 x) 2 log10 x − 2 log10 x

=1

1 – 8 log102x = log10x – 2 log102x

Let log x = t

log10x = 0 x = 10°

0 = 6log10 2 x + log10x – 1 Mathematics Online: It’s all about Mathematics

A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics 0 = (3 log10x – 1) (2 log10x + 1) = 0 ∴ 3 log10x = 1 or 2 log10x = – 1 log10x = x= 174.

3

1 3

log10x = – x=

10

(4.3x

(9x

log2 – 6) – log2 – 6) = 1 log2(4.t – 6) – log2 (x2 – 6) = 1 log2

4t − 6 t2 − 6

A. I. Prilepko Solution

1 2

1 10

Let 3x = t

=1

4t – 6 = 2(t2 – 6) 0 = 2t2 – 12 – 4t + 6 0 = 2t2 – 4t – 6 0 = t2 – 2t – 3 t2 – 3t + t – 3 = 0 ∴ t (t – 3) + 1(t – 3) = 0 (t + 1) (t – 3) = 0 ∴ t = 3, – 1 3x = 3 x=1 3x = – 1 Rejected ∴ Answer = 1

175.

1 log (5x – 4) + log 2

0.18 x + 1 = 2 + log 100

1 1 log10 (5x – 4) + log10x + 1 = 2 + log1018 – log10100 2 2 1 log10 (5x – 4) (x – 1) = log1018 2 5x2 – 4x + 5x – 4 = 324 5x2 + x – 328 = 0 2 D = b – 4ac = 1 + 6560 = 6561

⇒

D = 6561

x≡– ∴ 176.

82 ,8 10

Final Ans. 8

log4 (2.4x – 2 – 1) + 4 = 2x

 21x  log22  2. 2 − 1 + 4 = 2x  4   2.2 2x − 16  1   log2   + 4 = 2x 16 2  

 2 2x +1 − 16    log2   + 4 = 4x 16  

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A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics

A. I. Prilepko Solution

2 2 x +1 − 16 16

177.

logx 5 + logx 5x – 2.25 = (logx 5 )2 logx

 11  5 + logx5 + logxx – 2.25 =  2 log x 50   

2

1 1 1 + + 1 – 2.25 = 2 log5 x log5 x 4 log5 x

Let log5x = t

1 1 1 + + 1 – 2.25 = 2t t 4t

=

1 + 2 + 2t − 2t(2.25 ) 1 = 2t 4t

2 + 4 + 4t – 4t (2.25) = 1 6 + 4 (t – 2.25 t) = 1 – 6 –1 – 1.75 t = 178.

−5 5 = 4 7

log (log x) + log (log x4 – 3) = 0 log10 [log x + 4 logx – 3] = 0 5 log x – 3 = 1 5 log10x = 4 log10x =

4 5

Let logx = t

x = 104/5 ⇒ 5 log2x – 3 log ? = 0 log10t + 4 log2t – 4 log 3 179.

180.

log3x – 2 log1/3 x = 6 log3x + 2log3x = 6 log3x + log3x2 = 6 x3 = 36 x3 = (32)3 x=9 2 log x log(5 x − 4) = 1

⇒

2 log x = log (5x – 4) log x2 = log (5x – 4) x2 – 5x + 4 = 0 x2 – 4x – x + 4 = 0 x(x – 4) –1(x – 4) = 0 (x – 1) (x – 2) = 0 x = 1, 4 Answer = 4

181.

2 log82x + log8x2 + 1 – 2x = log8(2x)2 + log8 x2 – 2x + 1 =

4 3

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Mathematics Online: It’s all about Mathematics log84x2 (x2 – 2x + 1) =

A. I. Prilepko Solution

4 3

4x2(x2 – 2x + 1) = (23) 4x2(x2 – 2x + 1) = 24 4x2 (x2 – 2x + 1) = 16 x2 [x2 + 2x + 1] = 4 x4 + 2x3 + x2 – 4 = 0 2 log82x + log8(x2 + 1 – 2x) =

4 3

2[log82 + log8x] + log8 x2 – 2x + 1 =

4 3

(x + 1) (x3 – 3x2 + 4x – 4) (x – 2) (x2 – x + 2) Thus (x – 1) (x – 2) (x2 – x + 2) are its factor since 1 can’t satisfies, 2 satisfies the answer ∴ Answer = 2

182.

1 1 log2(x – 2) – = log1/8 6 3

3x − 5

1 1 log2 (x – 2) = = log2−3 6 3

3x − 5

1 1 1 log2 (x – 2) – =– log2(3x – 5)1/2 6 3 3 1 1 log2 (x – 2)1/6 . (3x – 5)1/6 = 6 3

= log2 (x – 2)1/6 . (3x – 5)1/6 =

1 3

((x – 2)(3x – 5))1/6 = (2)1/3 × 6 if 9 multiplied power by, get 3x2 – 6x –5x + 10 = 4 3x2 – 11x + 6 = 0 3x (x – 3) – 2(x – 3) = 0 x=

2 ,3 3

Since ∴ 183.

2 doesn’t satisfic equation 3

final answer = 3

2 log3(x – 2) + log3 (x – 4)2 = 0 log3(x – 2)2 + log3 (x – 4)2 = 0 2 log3(x – 2) + 2 log3 (x – 4) = 0 2[log3(x – 2) (x – 4)] = 0 x2 – 2x – 4x + 8 = 3° x2 – 6x + 8 – 1 = 0 x2 – 6x + 7 = 0

D = b2 – 4ax 36 – 28 = 8 D =2 2

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A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics ∴

x=

6±2 2 2

=2

(3 ± 2 ) 2

=3+ 2,3– 184.

A. I. Prilepko Solution

2

log2 (2 x 2 ). log 4 16 x = log4x3

let log2x = t

(log2 2 + log 2 x 2 )(log4 16 + log 4 x ) = 3 log4x

⇒

1   (1 + 2 log 2 x ) 2 + log2 x  = 3 log x 2 2   2

   (1 + 2t )  2 + t   =  2    

2

3  =  t 2 

2

2 4+t  = 9t (1 + 2t)   2  4

2 (4 + 8t + t + 2t2) = 9t2 0 = 9t2 – 4t2 – 18t – 8 0 = 5t2 – 18t – 8 0 = 5t2 – 20t + 2t – 8 0 = 5t (t – 4) + 2(t – 4) 0 = (5t + 2) (t – 4) t = 4, –

∴

log2x = 4

x = 24 x = 16

185.

186.

2 5

∴

log2x = –

x = (2)–2/5

2 5

can’t satisfied me

3 log x + 19 3 log x − 1 = 2 logx + 1

Let log10x = t

3 t + 19 = 2t + 1 3t − 1

log10x = 2

3t + 19 = (2t + 1)(3t – 1)

log10x = –

3t + 19 = 6t2 + 3t – 2t –1 0 = 6t2 – 2t – 20 0 = 3t2 – t – 10 0 = 3t2 – 6t + 5t – 10 0 = (3t + 5) (9t – 2)

x = 10– 5/3

log( x + 1 + 1) log 3 x − 40

5 3

=3

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A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics

A. I. Prilepko Solution

 3  log ( x + 1 + 1) = 3  log x − 40    x + 1 + 1 = logx – 120

log

120 = logx – log

x +1 – 1

187.

log326 – log32 2 = (log2x – 2) log3121 log32 2 + log323 – log322 = (log2x – 2) log312 (log3x)2 = (log102x – 2) (log33 + log34)

188.

1–

1 1 log (2x – 1) = log (x – 9) 2 2

1=

1 1 log (x – 9) + log (2x – 1) 2 2

1 [log (x – 9) (2x – 1)] 2 2 = log10 2x2 – 19x + 9 102 = 2x2 – 19x + 9 0 = 2x2 – 19x – 91 0 = 2x2 – 26x + 7x – 91 0 = 2x (x – 13) + 7(x – 13) 0 = (2x + 7) (x – 13)

1=

7 , 13 2 x = 13 Ans.

x=–

189.

log x + 7 − log 2 =–1 log 8 − log( x − 5)

log10

x+7 2 8

log10 x −5

log10 190.

=–1

8 x+7 = ? – log10 x −5 2

log(3x2 + 7) = log (3x – 2) = 1 log (3x2 + 7) = 1 log (3x – 2) = 1 101 = 3x2 + 7 101 = 3x – 2 3 = 3x2 12 = 3x x=1 x=4 log (3x2 + 7) – log (3x – 2) = 0 log

3x 2 + 7 = 0 3x − 2

3x2 + 7 = 3x – 2 3x2 – 3x + 9 = 0 x2 – x + 3 = 0

191.

1 log(x2 – 16x + 20) – log 3

3

1 7 = 3 log (8 – x)

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A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics

A. I. Prilepko Solution

1 1 [log10x2 – 16x + 20 – log107] = log108 – x 3 3

log10

x 2 − 16x + 20 = log108 – x 7

x2 – 16x + 20 = 56 – 7x x2 – 16x + 20 – 56 + 7x = 0 x2 – 9x – 36 = 0 – 12 + 3 x(x – 12) + 3(x – 12) = 0 (x + 3) (x – 12) = 0 ∴ x = –3, 12 makes the x < 1 rejected x = –3 Ans. 192.

log63x + 3 – log6 3x – 2 = x log6 2x + 3 3x − 2 2x + 3 3x − 2

2x + 3 3x − 2

=x

= 6x

= 2x . 3x

2x 3x 9 2x × x = 2x .3x 3x = 3 8 32

193.

1   1 +  log 3 + log 2 = log (27 – 2x  

194.

log(5x – 2 + 1) = x + log 13 – 2 log 5 + (1 – x) log 2

196.

log2(4x + 1) = x log2(2x + 3 – 6)

2

3)

 2x    log2 (22x + 1) = x + log2  8 − 6    198.

log3 (9x + 9) = x + log3 (28 – 2 . 3x)

199.

log (log x) + log (log x3 – 2) = 0 log10 [log x + logx3 – 2] = 0 4 log x – 2 = 1 log x =

Pattern may be same of Q.No. 178

3 4

x = 103/4 Answer has been given is 110.

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A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics 200.

log log

5

4x − 6 5

2 log5

log5

4x – 6 = – log

2x − 2

5

A. I. Prilepko Solution

2x– 1 = 2 Let 2x = t

=2

t2 − 6 =2 t−2

2x = 4

t2 − 6 =1 t−2

2x = 4

t2 − 6 = 5a1 t−2

∴

x=6

x=2

t2 – 6 = 5t – 10 t2 – 5t + 4 = 0 t(t – 1) – 4(t – 1) = 0 t = 4, 1

203.

 x 24  log  3 − x 2 

204.

 1   5

205.

x

206.

log2 (25x + 3 – 1) = 2 + log2 (5 x + 3 + 1)

210.

log2(2x2) . log2 (16x) =

 1 log 4 x   = 2 + 4 log 16 – 2 

log2 x −log x

=

2 3 log2 x − log x 3

1 . 5logx – 1 125

3 = 100 10

9 log22x 2

(log22 + log2x2) (4 + log24) =

9 log22 x 2

9 log22x 2 2 [4 + 8 log2x + log2x + 2log22x] = 9 log22x 8 + 16 log2 x + 2 log2x + 4 log22x = 9 log22 x 0 = 5 log22x – 18 log2x – 8 Let log2x = t 0 = 5t2 – 18t – 8t D = b2 – 4ac = 324 + 160 = 484

(1 + 2 log2x) (4 + log2x) = ⇒ ⇒

D = 22

t= ∴

18 ± 22 40 = =4 10 10

log2x = 4

log2x = –

x = 24

x = (2)– 2/5

2 5

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A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics 16 = 2 – 2/5

211.

1    log3 = log  3 3 + 27  log44 + 1 + 2 x    since radical should be defined 1+

1 24

= log  x 3 + 27    log (4.3 . 31/2x) = log (31/x + 27) 12.t = t2 + 27 0 = t2 – 12t + 27 0 = t2 – 9t – 3t + 27 0 = (t – 9) (t – 3) ∴ t = 9, 3 log44 + log 3

Let 31/x = t 31/x = t

1 2

x = 1,

212.

A. I. Prilepko Solution

log(x3 + 27) – log (x2 + 6x + 9) = log log (x + 3) (x2 – 3x + 9) – log

3

7

( x 2 + 6x + 9)1/ 2 ( x + 3) 2

= log 7

 ( x + 3)( x 2 − 3 x + 9)    = log 7 10 ( x + 3)  

log10

x2 – 3x + 9 – 7 = 0 x2 – 3x + 2 = 0 x2 – 2x – x + 2 = 0 x(x – 2) – 1(x – 2) = 0 (x – 1) (x – 2) = 0 x = 1, 2 213.

215.

5log x = 50 – xlog 5 5log x + xlog 5 = 50 5log x + 5log x = 50 5t + 5t = 50 2.5t = 50 5t = 25 t=2

x

3 log x −

1 log x

=

3

Let log10x = t x = 102 = 100

10

2

x

3 log x −1 log x

log x

=

3

3 log2 x −1 log x

log x .

10 = log

3

10

3 log2 x − 1 1 = log 10 log x 3

3 log2x – 1 =

1 3

Mathematics Online: It’s all about Mathematics

A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics 3 log2x =

4 3

log102x =

4 9

log10x = ±

2 3

A. I. Prilepko Solution

x = (10)2/3 , (10–2/3 216.

|x – 10| log2 (x – 3) = 2(x – 10) log2 (x – 3)|x – 10| = 2(x – 10) (x + 10) log2(x – 3) = 2(x – 10)1 – (x + 10) log2(x – 3) = 2(x – 10) log2(x – 3) = – 2 log2(x – 3) = 2 x – 3 = 29 (x – 3) = 2–2 x=7

217.

(x – 3) =

1 4

⇒x=

1 13 +3 = 4 4

log4 log2x + log2 log4x = 2

log2x = t

1 1 log2 log2x + log2 log2x = 2 2 2

2x = 4

1 t log2t + log 2 =2 2 2 log2 t1/2 + log2t – log22 = 2 log2t1/2 + log2t – log22 = 2 log2t1/2 + log2t = 3

x = 24

1 log2t + log2t = 3 2 log2t + 2 log2t = 6 3 log2 t = 6 t = 22 = 4

218.

(6x – 5) |ln (2x + 2.3)| = 8 ln (2x + 2.3) 8ln (2 x + 2.3) (6x – 5) = | ln (2x + 2.3) |

Ist

6x – 5 = – 8 6x = 13 x=

IInd

13 1 =– 6 2

It is not in answer

6x – 5 = 8

The right answer

6x = 13

=

x=

13 −13 , 6 20

13 6

log9 9 x 8 − log3 3 x = log3x3

219. =

log9 9 + log9 x 8 .(log3 3 + log3 x ) = 3 log3x

Mathematics Online: It’s all about Mathematics

A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics

A. I. Prilepko Solution

2

  8 =  1 + log3 x.(1 + log3 x )  = (3 log3x)2 2   = (1 + 4 log3x) (1 + log3x) = 9 log32 x (1 + 4t) (1 + t) = 9t2 0 = 9t2 – 4t2 – 5t – 1 0 = 5t2 – 5t – 1 D = b2 – 4ac = 25 + 20 = 45 5±3 5 10

t=

log3x = t log3 x =

5±3 5 10

x=3

220.

5±3 5 10

log2 (100x) – log2 10x + log2x = 6 2 2 log102 102 + log102x – [log10 10 + log10 x] + log102x = 6 2 2 − 1 − log10 x =6 4 + log10 2 x= 3 log10

log10x = 221.

3

9 log1/ 3 ( x +1) = 5log1/ 5 ( 2 x

3

2 log

3 −1

( x +1)

= 5

log

5−1

2

+1)

( 2 x 2 +1)

x−7    x −1

−2

2 –1 3 −2 log3 ( x +1) = (2x + 1) (x + 1)–2 = (2x2 + 1)–1

1 ( x + 1)2

2x2

=

1 2

2x + 1

x2

+1= + 1 + 2x 2x2 – x2 – 2x = 0 x2 – 2x = 0 x(x – 2) = 0 x = 0, 2

223.

2 log2

log2 ⇒ ⇒ ⇒

 x − 1  =1 + log2   x + 1

x 2 + 49 − 14 x x −1 × =1 ( x − 1)( x − 1) x +1

x 2 + 490 − 14 x x2 − 1

=2

x2 + 49 – 14x = 2x2 – 2 x2 + 14x – 2 – 49 = 0 x2 + 14x – 51 = 0 x2 + 17x – 3x – 51 = 0

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A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics x(x + 17) – 3(x + 17) = 0 ∴ x = – 17, 3→can’t satisfy the equation. ∴ x = – 17 only

225.

A. I. Prilepko Solution

1 3 1 log (1 + x) + log (1 – x) = log (1 – x2) 2 2 2

=

1 1 [log (1 + x) + log (1 – x)3] = log (1 – x2) 2 2

log

⇒

1+ x (1 − x )2 (1 − x )

log

= log (1 – x2)

1+ x (1 − x )2 (1 − x ) 1+ x

log10

(1 − x )3 1− x2

(1 + x ) (1 − x )

log10

×

3

– log(1 – x2) = 0

=0

1 =1 (1 + x )(1 − x )

1 (1 − x )4

=0

log10 (1 – x)–4 = 0 – 4 log10 (1 – x) = 0 100 = 1 – x x=0

226.

log(35 − x 3 ) =3 log(5 − x ) log (35 – x3) = 3 log (5 – x) log (35 – x3) = log(5 – x)3 35 – x3 = 125 – x3 – 15x (5 – x) 35 – 125 = – 75x + 15x2 0 = 15x2 – 75 x + 90 0 = x2 – 5x + 6 0 = x2 – 2x – 3x + 6 ∴ x = 2, 3

227.

logx2 – log4x +

7 =0 6

1 7 1 log2 x – 2 log2x + 6 = 0

6 – 3 log22 x ± 7 log2x = 0 3 log22 x – 9 log2x + 2 log2x – 6 = 0 (3 log2x + 2) (log2x – 3) = 0 3 log2x = – 2 log2x = 3 log2x = – x = 2 – 2/3

2 3

x = 23 x=8

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A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics 229.

A. I. Prilepko Solution

log 2 x – 0.5 = log2 x log 2 x =

1 1 log2x + 2 2

1 [log2x + 1] 2 4 log2x = log22x + 1 + 2 log2x 0 = log22 x – 2 log2x + 1 0 = log2x (log2x – 1) – 1(log2x – 1) 0 = log2x – 1 1 = log2x x=2 log 2 x =

230.

log1/3

  1 x  2 ×   − 1 = log 1/3   2   x

 1  1 2   –1=   2 2

 1  2x     − 4   2  

2x

x

x

–4

Let

 1   =t 2

x

 1  1  1 2   –   .   =–3 2 2 2

⇒ ⇒ ⇒ ⇒ ⇒

2t – t2 = 3 0 = t2 – 2t – 3 0 = t(t + 1) – 3(t + 1) (t – 3) (t + 1) = 0 t = 3, –1

Q

 1   =t 2

x

x

 1   = – 1, 3 2

–1 not defined

x

 1 log   = log 3 2

log 2–x = log 3 – x log 2 = log 3 log 3 – x = log 2

x = – log23 Ans.

231.

1 1 log6(x – 2) + log6(x – 11) = 1 2 2 1 log6 (x – 2) (x – 11) = 1 2 log6 x2 – 13x + 22 = 2 x2 – 13x + 22 = 62 x2 – 13x + 22 – 36 = 0 x2 – 13x – 14 = 0 x2 – 14x + x – 14 = 0 x(x – 14) + 1(x – 14) = 0

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A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics (x + 1)(x – 14) = 0 x = 14 Ans. –1 is not defined 233.

A. I. Prilepko Solution

1/ 2 log7 2 + log72 x = log 7 −1 (3)

log72 +

1 1 log7x = – log73 2 2

1 1 log73 + log7x = 0 2 2 2 log72 + log7 3 + log7x = 0 log7 12 = – log7x x = – 12

log72 +

237.

log3x

3 + log32x = 1 x

3 x + log 2 x = 1 3 log3 3 x log3

Let log3x = t

log3x = 0

log3 3 − log3 x 2 log3 3 + log3 x + log 3 x = 1

log3x = 1

1− t + t2 = 1 1+ t

x=1

1 – t + t2 (1 + t) = 1 + t t3 + t2 – 2t = 0 t(t2 + t – 2) = 0 t2 + t – 2 = 0 t(t + 2) – 1(t + 2) = 0 t = 1, 2 Ans. 1, 3, 3–2 246.

x = 3–2

log4(x + 12) logx2 = 1 x + 12 1 log log x = 1 2 2

log2 (x + 12) = 2 log2x log2 x + 12 = log2x2 x + 12 = x2 0 = x2 – x – 12 0 = x2 – 4x + 3x – 12 0 = x(x – 4) + 3(x – 4) x = 4, 3 not satisfied ∴ 24 Ans. 247.

log16x + log4x + log2x = 7 1 1 log2x + log2x + log2x = 7 4 2 log2 x + 2 log x + 4 log2 x =7 4 7 log2x = 28 log2x = 4

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A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics x = 16

248.

log 10 +

1 log (3 2 3

1 2 log (3 3

log10 (3 2

x

x

x

+ 271) = 2

+ 271) = 1

+ 271) = 3

32

x

+ 271 = 103

32

x

= 1000 – 271

32

x

= 279

32

x

= 36

x =3

⇒

2 x =6

⇒

( x )2 = 9

250.

log5 (3x + 10) + 7.10 = log5 (9x + 156) Detective Question for me and how can I solve

251.

(log4x – 2) log4x =

3 (log4x – 1) 2 2(log4x – 2) log4x = 3 log4x – 3 (2t – 4) t = 3t – 3 2t2 – 4t – 3t + 3 = 0 2t2 – 7t + 3 = 0 2t2 – 6t + t + 3 = 0 2t (t – 3) – 1(t – 3) = 0

t = 3,

A. I. Prilepko Solution

⇒

x=9

Let log4x = t

1 2

∴

log4x = 3,

1 2

x = 43 , (4)1/2 x = 64, 2 252.

log5x + log25x = log1/5

3

1 1 log5x = – log5 3 2 2 2 log5x + log5x = – log 53 – 3 log5x = log53 log5x–3 = log5 3 (x–3)1/3 = (3)– 1/3

log5x +

x= 254.

1 3

3

log2 (9 – 2x) = 3 – x 23 – x = 9 – 2x

Let 2x = t

Mathematics Online: It’s all about Mathematics

A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics 23 2x

= 9 – 2x

2x = 1

8 =9–t t

2x = 8

8 = 9t – t2 t2 – 9t + 8 = 0 t2 – 8t – t + 8 = 0 t (t – 8) – 1 (t – 8) = 0 (t – 1) (t – 8) = 0 t = 1, 8 x = 0, x = 3 Ans. 255.

256.

2(log 2 – 1) + log (5

x

2(log 2 – 1) + log (5

x

logx3 + log3x = log

A. I. Prilepko Solution

x

x=3

+ 1) + log (51−

x

+ 5)

 5  + 5  + 1) + log  5 x 

3 + log3 x +

1 2

1 2 1 1 log3 x + log3x = log3 x + 2 log3x + 2 1 2 t 1 +t= + + t t 2 2

1+ t 2 2 + t2 + t = t 2t

2 + 2t2 = 2 + t2 + t t2 – t = 0 t = 0, 1

259.

Q

log3x = 0 x = 30 = 1 log3x = 1

logx 125x . log225 x = 1

(log5 2 x )2

log5 125 x 2 log5 x . log5 2x = 1

1   log5 x  2 

⇒

x=3

2

log5 53 + log5 x 1 . log25x = 1 log5 x 2 3+t t2 . =1 t 4

3t 2 + t 3 =1 4t

t(3t + t2) = 4t t2 + 3t – 4 = 0 t2 + 4t – t – 4 = 0 t(t + 4) – 1(t + 4) = 0 (t – 1) (t + 4) = 0 t = 1, – 4 ∴ log5x = 1, –4 = 5, 5–4It’s all about Mathematics MathematicsxOnline:

A. I. Prilepko Solution