Mathematics Tricks & Tips of logarithm
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Solving logarithmic equations using log properties

Q 2107012888

If `log_(a) (ab) = x,` then what is `log
_(b) (ab)` equal to?
NDA Paper 1 2016
(A)

` 1/x`

(B)

`x/ (x+1)`

(C)

`x/ (1-x)`

(D)

`x/ (x-1)`

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)`
Correct Answer is `=>` (D) `x/ (x-1)`
Q 1658423304

If `log_(10) 2, log_(10) (2^x -1)` and `log_(10) (2^x + 3)` are three

consecutive terms of an `AP`, then the value of `x` is
NDA Paper 1 2015
(A)

`1`

(B)

`log _(5) 2`

(C)

`log _(2) 5`

(D)

`log _(10) 5`

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`
Correct Answer is `=>` (C) `log _(2) 5`
Q 2241223123

If `log_8 m + log_ 8 (1/6) = 2/3 `.then `m` is equal to
NDA Paper 1 2015
(A)

`24`

(B)

`18`

(C)

`12`

(D)

`4`

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`
Correct Answer is `=>` (A) `24`
Q 2309856718

What is the value of `log_2 (log_3 81)`?
NDA Paper 1 2011
(A)

`2`

(B)

`3`

(C)

`4`

(D)

`9`

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)`
Correct Answer is `=>` (A) `2`
Q 2349056813

What is `log (a + sqrt(a^2 + 1)) + log ( 1/(a + sqrt(a^2 + 1) ))`
equal to?
NDA Paper 1 2011
(A)

`1`

(B)

`0`

(C)

`2`

(D)

`1/2`

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`
Correct Answer is `=>` (B) `0`
Q 2319056819

What is the value of
`log_(10) (9/8) - log_(10) ((27)/(32)) + log_(10) (3/4)`?
NDA Paper 1 2011
(A)

3

(B)

2

(C)

1

(D)

0

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`
Correct Answer is `=>` (D) 0
Q 2339156912

If `log_3 [log_3 [log_3 x]] = log_3 3`, then what is the
value of `x`?
NDA Paper 1 2010
(A)

`3`

(B)

`27`

(C)

`3^9`

(D)

`3^(27)`

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)`
Correct Answer is `=>` (D) `3^(27)`
Q 2359167014

What is the value of `(log_sqrt(alpha beta ) (H)) /(log _sqrt(alpha beta gamma) (H))`?

NDA Paper 1 2010
(A)

`log_(alpha beta ) (alpha)`

(B)

`log_(alpha beta gamma) (alpha beta)`

(C)

`log_(alpha beta) (alpha beta gamma )`

(D)

`log_(alpha beta) ( beta)`

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`
Correct Answer is `=>` (C) `log_(alpha beta) (alpha beta gamma )`
Q 2339267112

What is the value of `((log_(27) 9)(log_(16) 64))/(log_4 sqrt(2))`
NDA Paper 1 2009
(A)

1

(B)

2

(C)

4

(D)

8

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`
Correct Answer is `=>` (C) 4
Q 2329367211

If `(log_x x)(log_3 2x)(log_(2x) y) = log_x x^2`
, then what is the value of `y`?
NDA Paper 1 2009
(A)

`9/2`

(B)

`9`

(C)

`18`

(D)

`27`

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`
Correct Answer is `=>` (B) `9`
Q 2369367215

If `log_k xx log_5 k = 3`, then what is `x` equal to?
NDA Paper 1 2009
(A)

`k^5`

(B)

`5k^3`

(C)

`243`

(D)

`125`

Solution:

Given, `log_5 k log_k x = 3`.

`=> log_5 x = 3`

`:. x = 5^3 = 125`
Correct Answer is `=>` (D) `125`
Q 2379467316

If `x > 1` and `log_2 x, log_3 x, log_x 16` are in GP, then
what is `x` equal to?
NDA Paper 1 2009
(A)

`9`

(B)

`8`

(C)

`4`

(D)

`2`

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`
Correct Answer is `=>` (A) `9`
Q 2319556419

If `log_(10) (x + 1) + 1og_(10) 5 =3,`then what is the
value of `x`?
NDA Paper 1 2009
(A)

`199`

(B)

`200`

(C)

`299`

(D)

`300`

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`
Correct Answer is `=>` (A) `199`
Q 2309656518

If ` 10^ (log_(10) | x|) = 2`, then what is the value of `x`?
NDA Paper 1 2008
(A)

Only 2

(B)

Only -2

(C)

2 or -2

(D)

1 or -1

Solution:

`∵ 10^(log 10 | x | ) = 2`

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

`:. x = 2` or `-2`
Correct Answer is `=>` (C) 2 or -2
Q 2379756616

For what value (s) of `x` is
`log_(10) (999 + sqrt(x^2 - 3x + 3)) =3` ?
NDA Paper 1 2007
(A)

`0`

(B)

Only 1

(C)

Only 2

(D)

`1 , 2`

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`
Correct Answer is `=>` (D) `1 , 2`
Q 2339856712

Find the value of `log_(5sqrt5) 5`.
NDA Paper 1 2007
(A)

`2/3`

(B)

`1/3`

(C)

`1/2`

(D)

`2`

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`
Correct Answer is `=>` (A) `2/3`


 
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