Logarithmic Function MCQ Quiz in मराठी - Objective Question with Answer for Logarithmic Function - मोफत PDF डाउनलोड करा

Given:
log 2 = 0.3010

log 3 = 0.4771

Formula used:

log (x × y) = log x + log y

Calculation:

log 6 = log (2.3)

⇒ log (2.3) = log 2 + log 3

⇒ log (2.3) = 0.3010 + 0.4771 = 0.7781

∴ log 6 = 0.7781.

Concept:

Calculation:

Hence, option (2) is correct.

log 8 = 0.903

log (2)³ = 0.903                         [∵ log a^b  = b log a]

log 2 = 0.903/3 = 0.301

Concept:

Logarithm properties

Product rule: The log of a product equals the sum of two logs.

Quotient rule: The log of a quotient equals the difference of two logs.

Calculation:

Given:

log (x + 2) + log (x − 2) = log 5

⇒ log [(x + 2) (x – 2)] = log 5                (∵ )

⇒ log (x ² – 4) = log 5

⇒ x² – 4 = 5

⇒ x² = 9

∴ x = ± 3

Here x ≠ -3 is not possible because (x – 2) should be greater than zero.

∴ x = 3

Given:

We have a logarithm expression 'Log327'

Concept:

Logarithm

Formula Used:

Logab = x 

ax = b

Calculation:

Log327 = n

⇒ 3n = 27

⇒ 3n = 33

⇒ n = 3
∴ The value of Log327 is 3.

Concept Used:-

The base rule of logarithm for a, b and x is given as,

0\)

Here, a is a positive integer.

Also, the power rule of logarithm is given as,

Explanation:-

Given,

log0.3 (x – 1) 0.09 (x – 1),

With the help of above log rule, it can be written as,

On comparing both sides, we get,

\sqrt{x-1}\ \ \ \text{ and}\\ \Rightarrow \sqrt{x-1}>0\)

Now we have,

\sqrt{x-1} \\ & \Rightarrow (\sqrt{x-1})^20 \\ & \Rightarrow (x-2)(x-1)>0 \\ & \Rightarrow x2 \end{aligned}\)

Also.

With the obtained intervals, we get,

x > 2

So the value of x will lie in the interval (2, ∞).

Correct option is 1.

Given:

log_x 4 + log_x 16 + log_x 64 = 12

Formula Used:

If log_x y = a, then x^a = y

log_x a^b = b log_x a

Calculation:

log_x 4 + log_x 16 + log_x 64 = 12

⇒ log_x 2² + log_x 2⁴ + log_x 2⁶ = 12

⇒ 2 log_x 2 + 4 log_x 2 + 6 log_x 2 = 12

⇒ 12 log_x 2 = 12

⇒ log_x 2 = 1

⇒ 2 = x¹

∴ x = 2

Given:

The value of [log10(5 log10 100)]²

Concept used:

log a^m = m log a

log_a a = 1

Calculation:

 [log10(5 log10 100)]²

⇒  [log10(5 log10 10²)]²

⇒  [log10(10 log10 10)]²    [As log am = m log a]

⇒ log₁₀(10 × 1)]²   [As loga a = 1]

⇒ [log₁₀ 10]²

⇒ [1]² = 1

∴ The require value is 1.

Concept:

Calculation:

We know that

Let y = tan h^-1 (x)

⇒ x = tan hy

⇒ xe^2y + x = e^2y - 1

⇒ e2y(1 - x) = 1 + x

On taking log both sides, we get,

Given:
log 2 = 0.3010

log 3 = 0.4771

Formula used:

log (x × y) = log x + log y

Calculation:

log 6 = log (2.3)

⇒ log (2.3) = log 2 + log 3

⇒ log (2.3) = 0.3010 + 0.4771 = 0.7781

∴ log 6 = 0.7781.