Aptitude Logarithm Test Paper 4

Answer: B

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 q^p

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

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

Or, log x = log (x^p y^q z^r)

Since, both sides log will be canceled out

Therefore, x = x^p y^q z^r)

Hence, the antilogarithm of pa+qb+rc = x x^p y^q z^r)

Answer: C

Explanation:

We have an expression log_b a² log_c b² log_a c²
Now, we know that p log q = log q^p

Similarly, (2 log_b a) (2 log_c b) (2 log_a c)
Now, [2*2*2]

Therefore, 8 * 1 = 8

Answer: A

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

Answer: A

Explanation:

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

We know that, logx y = a, is equals to x^a = 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.

Answer: C

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