Solution:

Given,`log_(a) (ab) = x`

` => log_(a) a + log_(a) b = x`

` => 1 + log _(a) b = x`

` log _(a) b = x - 1`

Now, `log_(b) ab = log_(b) a + log_(b) b`

` = 1/ (log_(a) b) +1 = 1/ (x -1) +1`

` = ( 1 +x -1)/(x-1) = x/ (x-1)`

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Solution:

If `a, b` and `c` are in `AP`, then `2b = a + c`

`:. 2log_(10) (2^x - 1) = log_(10) (2^x + 3) + log_(10) 2`

`=> log_(10) (2^x - 1)^2 = log_(10) (2^x + 3) xx 2`

` => (2^x - 1)^2 = 2^(x + 1) + 6`

`=> (2^x)^2 + 1 - 2 . 2^x =2^(x + 1) + 6`

`=> 2^(2x) - 2^(x+1) -2^(x+1) = 6-1 => 2^(2x) - 2 . 2 ^(x+1) - 5 = 0`

Let `2^x = y`, then

`y^2 - 4y- 5 = 0`

`=> y^2 - 5y + y- 5 = 0`

`=> y (y - 5) + 1 (y - 5) = 0`

`=> (y+ 1)(y- 5)= 0`

`=> y = -1` or `y = 5`

`=> 2^x = - 1` [not possible]

or `2^x = 5`

`=> log2^x = log 5 => xlog^2 log 5`

` x = (log5)/(log2) = log_2 5`

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Solution:

We have, `log_8 m + log_ 8 (1/6) = 2/3 `

` => log_8 (m/6) = 2/3 ``\ \ \ \ \[∵ log _(a) m+ log _(a) n = log a (m xx n)]`

` => m/6 = (8)^(2//3) = (2^3)^(2//3) = 2^2 = 4`

` => m = 6 xx 4 = 24`

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Solution:

Given, `log_2 (log_3 81)`

`=log_2 {log_3 (3)^4}= log_2 { 4 1og _3 3}`

`= log_2 (4) ( ∵ log_3 3 = 1)`

`= log_2 (2)^2 = 2 log_2 2`

`= 2 · 1 = 2( ∵ log_2 2 = 1)`

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Solution:

`log (a + sqrt(a^2 + 1 )) + log ( 1/(a + sqrt(a^2 + 1) ))`

` = log (a + sqrt(a^2 + 1)) + log (a + sqrt( a ^2 + 1))^(-1)`

`= log (a + sqrt(a^2 + 1)) - log (a + sqrt(a^2 + 1))`

` = 0`

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Solution:

`log_(10) (9/8) - log_(10) ((27)/(32)) + log_(10) (3/4)`

(` ∵ log m + log n = log mn` and `log m - log n =log m//n`)

`= (log_(10)) 9/8 . (32)/(27) . 3/4 = log_(10) 1 = 0`

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Solution:

`log_3 [log_3 [log_3 x]] = log_3 3`

`=> log_3 [log_3 x] = 3 ( ∵ log_a x = bx = a^b)`

`=> log 3^x = 3^3`

`=> log_3 x = 27 => x = 3^(27)`

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Solution:

`(log_sqrt(alpha beta ) (H)) /(log _sqrt(alpha beta gamma) (H)) ( ∵ log_a b = 1/(log_b a))`

`= log_sqrt(alpha beta) sqrt(alpha beta gamma)`

`= 1/2 1og_sqrt(alpha beta) (alpha beta gamma) = (1//2)/(1//2) log _(alpha beta) (alpha beta gamma)`

`(∵ log_(a^m) b = 1/m log_a b)`

`= log_(alpha beta) alpha beta gamma`

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Solution:

`((log_(27) 9)(log_(16) 64))/(log_4 sqrt(2))`

( ` ∵ log_(a^n) b = 1/n log_(a^ b)` and `log_x x = 1`)

` = (log_(3^3) (3^2) log_(4^2) (4)^3)/(log_(2^2) (2^(1//2)))`

`= ( 2/3 log_3 3·3/2 log_4 4 )/(1 /(2.2) log_2 2) = 1/(1/4) = 4`

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Solution:

`(log_x x) (log_3 2x)(log_(2x) y) = log_x x^2`

`=> 1 (log_3 2x)(log_(2x) y) = 2 ( ∵ log_x x = 1)`

`=> log_3 y = 2 ( ∵ log_a x · log_x b = log_a b)`

`:. y = 3^2 = 9`

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Solution:

Given, `log_5 k log_k x = 3`.

`=> log_5 x = 3`

`:. x = 5^3 = 125`

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Solution:

Since, `log_2 x,log_3 x,log_x 16` are in GP.

`:. (log_3 x)^2 =1og_2 x· log_x 16`

`=> (log_3 x)^2 = log_2 16`

`( ∵ log_a x ·log_x b = log_a b` and `log_x x = 1)`

`=> (log_3 x )^2 = 4 log_2 2`

`=> log_3 x = 2`

`:. x = 3^2 = 9`

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Solution:

Given. `log_(10) (x + 1) + log_(10) 5 = 3`

`=> log_(10) 5(x + 1) = 3`

`=> 5(x + 1 ) = (10)^3`

`=> 5x+5 = 1000`

` => 5x = 995`

` :. x = 199`

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Solution:

`∵ 10^(log 10 | x | ) = 2`

`=> log_(10) = | x | = log _(10)2 => | x | = 2`

`:. x = 2` or `-2`

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Solution:

`log_(10) (999 + sqrt(x^2 - 3x + 3)) =3`

`=> 999 + sqrt( x^2 - 3x + 3) = 1000 => sqrt( x^2 - 3x + 3 ) = 1`

`=> x^2 - 3x + 3 = 1 => x^2 - 3x + 2 = 0`

`=> x^2 - 2x - x + 2 = 0 => x(x - 2)- 1(x - 2) = 0`

` => (x - 1)(x - 2) = 0`

` :. = 1, 2`

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Solution:

`log_(5sqrt5) 5 = ( log 5)/( log 5 sqrt5) = ( log 5)/( log 5 + 1/2 log 5) `

` = ( log 5)/(3/2 log 5 ) = 2/3`

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