# Aptitude Logarithm Test Paper 4

16) If log10 x = a, log10 y = b, and log10 z = c, so (to base 10) the antilogarithm of pa+qb+rc would be

1. Px + qy- rz
2. xp yq zr
3.

Explanation:

Let x = antilog (pa+qb+rc)z

Or, we can say that log x = pa+qb+rcz

Or, log x = p log10 x + q log10 y + r log10 z

We know that, p log q = log qp

Or, log x = log10 xp + log10 yq + log10 zr

Or, we know that log a + log b = log ab

Or, log x = log (xp yq zr)

Since, both sides log will be canceled out

Therefore, x = xp yq zr)

Hence, the antilogarithm of pa+qb+rc = x xp yq zr)

17) The value of logb a2 logc b2 loga c2 is

1. 1
2. 7
3. 8
4. 9

Explanation:

We have an expression logb a2 logc b2 loga c2
Now, we know that p log q = log qp

Similarly, (2 logb a) (2 logc b) (2 loga c)
Now, [2*2*2]

Therefore, 8 * 1 = 8

18) If log(x2 - 4x + 5) = 0, what will be the value of x?

1. 2
2. 3
3. 0
4. 1

Explanation:

We have log(x2 - 4x + 5) = 0

Note: when log moves from one side to the other, it will become exponential.

So, (x2 - 4x + 5) = e0

Or, we know that exponential to the power zero is one.

So, x2 - 4x + 5 = 1

Now, x2 - 4x + 5-1 = 0

Or, x2 - 4x + 4 = 0

Or, (x-2) 2 = 0

Therefore, x- 2 = 0

Hence, x = 2

19) If log2 [log3 (log2 x)] = 1, what is the value of x?

1. 512
2. 1
3. 0
4. 128

Explanation:

We have an expression log2 [log3 (log2 x)] = 1

We know that, logx y = a, is equals to xa = y

Similarly, [log3 (log2 x)] = 21

Now, log2 x = 32 = 9

Now, x = 29 = 512

Therefore, the value of x from the expression log2 [log3 (log2 x)] = 1, is 512.

20) is equal to

1. 1
2. 2
3. 0
4. 3

Explanation:
We have an expression

We know that log m + log n + log p = log (m * n * p)
Similarly, log

Now, we can write it as, log => log (1)

Note: logx (1) is always equals to be zero.
Therefore, is equals to zero.

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