Aptitude Logarithm Test Paper 3

Answer: C

Explanation:
Log₁₀ 2 + 16 log₁₀ + 12 log₁₀ + 7 log₁₀

We know that log mⁿ = n log m

So, Log₁₀ 2 + log₁₀ [2⁴/ 3*5] ¹⁶ + log₁₀ [5²/ 2³ * 3] ¹² + log₁₀ [3⁴/ 2⁴ * 5] ⁷

Or, log₁₀ 2 + log₁₀ [2⁶⁴/ 3¹⁶ * 5¹⁶] + log₁₀ [5³⁶/ 2³⁶ * 3¹²] + log₁₀ [3²⁸/ 2²⁸ * 5⁷]

We know that log m*n = log m + log n

Now, log₁₀ [2 * 2⁶⁴ * 5²⁴ * 3²⁸]/ [3¹⁶ * 5¹⁶ * 2³⁶ * 3¹² * 2²⁸ * 5⁷]

Or, log₁₀ [2⁶⁵ * 3²⁸ *5²⁴]/ [2⁶⁴ * 3²⁸ *5²³]

Therefore, log₁₀ (2*5) = log₁₀ (10) = 1

Answer: A

Explanation:

We have log (11 + 4√7) = log (2 + x)

Now, we can write it as log (7 + 4 + 4√7) = log (2 + x)

We know that (a+b) ² = a² + b² + 2ab

Similarly, log (2 + √7) ² = log(2 + x)

Or, log (2 +√7) = log (2 + x)

Both side Log will be canceled out

Now, 2 + √7 = 2 + x

Therefore, x = 2 + √7 - 2 = √7

Answer: A

Explanation:

We have log₁₀ = 2

Or, we can write it as log₁₀ = 2 log₁₀ 10

Or, log₁₀ = 2 log₁₀ 100

Therefore, both side log will be canceled out.

Or, = 4

We know that √x = a, or x^1/2 = a, then x = a²

Similarly, x² - 12x + 36 = 4²

Or, x² - 12x + 36 = 16

Or, x² - 12x + 20=0

Now, (x-2)(x-10) = 0

Or, x-2 = 0, and x-10 =0

Therefore, we can say that x = 2, x = 10

Answer: C

Explanation:
We have an expression f(a) = log , then we have to find the value of f(2a / 1+a²).

We can find it by replacing a with

Now, f= log

Or, log

Now, the denominator of both fractions will cancel out each other.
Now, f= log

Or, log = log

We know that log mⁿ = n log m.
Therefore, 2 log is equals to 2 f(a).

Answer: D

Explanation:
We know that, if a number is in the form

Then the output will be the number itself, i.e., the expressionis equals to 10.

Since, log_m m = 1

Therefore, log₁₀ (10) = 1