# Aptitude Logarithm Test Paper 1

1) Find the logarithm of 1/256 to the base 2√2.

1. 16
2. 13/5
3. -16/3
4. 12

Explanation:

Let log2√2 [1/256] = x

We know that loga y = x is similar to ax = y

So, we can write it as [1/256] = (2√2) x

Or, (2√2) x = [1/28]

Or, [21 * 21/2]x = 1/28

Or, 23x/2 = 2-8

Therefore, 3x/2 = -8

Hence, x = (-8 * 2)/ 3 = -16/3

2) If loga [1/36] = -2/3, find the value of a.

1. 6
2. 8
3. 9
4. 216

Explanation:

ATQ, loga [1/36] = -2/3

We know that loga y = x is similar to ax = y

So, a (-2/3) = 1/36

Or, a (-2/3) = 1/62

Or, a (-2/3) = 6-2

Now, multiply and divide by 3 in the power of 6 to make the power equals to power of a.

So, a (-2/3) = 63(-2/3)

On comparing both side, we get a = 63

Therefore, a = 216

3) Find the value of x

Log4(log8 64) = log5 x

1. 2
2. 6
3. 5
4. √5

Explanation:

ATQ, Log4 (log8 64) = log5 x...... (i)

Let, log8 64 = a

Or, 64 = 8a, or 8a = 82

That means a=2

Now, put log8 64 = 2in equation 1.

Log4 (2) = log5 x...... (ii)

Now, let log4 2 = s

Or, 4s = 2, or 22S = 2

Now, on comparing both side we get 2s = 1, or s = ½

Put the value of s in equation 2

So, log5 x = ½

Therefore, x = 51/2 = √5

4) The equation, a2 + b2 = 7ab equals to Explanation:

Here, a2 + b2 = 7ab

Add both sides 2ab to make a formula.

Or, a2 + b2 + 2ab = 7ab + 2ab

Or, (a+b) 2 = 9ab

Or, (a+b) 2 / 9 = ab

Or, [1/3 (a+b)]2 = ab

Now, taking log both sides

Therefore, log [1/3 (a+b)]2 = log ab

We know that log m*n = log m + log n, and log mn = n log m

So, 2 log [1/3 (a+b)] = log a + log b

Therefore, log [1/3 (a+b)] = 1/2(log a + log b)

= a2 + b2

= 7ab

5) If (log3 x)(logx 2x)(log2x y) = logx x2, find y.

1. 6
2. 9
3. 5
4. 7

Explanation:

(log3 x)(logx 2x)(log2x y) = logx x2
We know that logb a = Now, A = B = C = Now, ATQ, A* B * C = logx x2
Or, * * = We know that, loga mn = n loga m = We know that loga a = 1
That means, log3 y = 2
We know thatloga y = x is similar to ax = y
Therefore, y = 32 = 9

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