ₓ arrow 1}≤ft( ₃ 3 x)^{ ₓ 8}=

Correct Answer is : elog⁡38e^{ ₃ 8}elog38

Correct Answer is : elog⁡38e^{ ₃ 8}elog38

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MHTCET 22 Apr 2025 Shift 2 Paper

Question : 111 of 150

Marks: +1, -0

\lim _{x \rightarrow 1}\left(\log _{3} 3 x\right)^{\log _{x} 8}=\ldots

Solution:

Concept:

Use logarithm properties to rewrite the base and exponent, then apply the standard limit

\lim_{u \to 0}(1+u)^{1/u}=e

Explanation:

Let

L = \lim_{x \to 1}\left(\log_3 3x\right)^{\log_x 8}

Write

\log_3 3x = \frac{\log(3x)}{\log 3} = \frac{\log 3 + \log x}{\log 3} = 1 + \frac{\log x}{\log 3}

Also,

\log_x 8 = \frac{\log 8}{\log x}

Therefore,

L = \lim_{x \to 1}\left(1 + \frac{\log x}{\log 3}\right)^{\frac{\log 8}{\log x}}

x \to 1

\log x \to 0

. Put

u = \frac{\log x}{\log 3}

, so

u \to 0

Then

\frac{\log 8}{\log x} = \frac{\log 8}{u \log 3}

Thus,

L = \lim_{u \to 0}(1 + u)^{\frac{\log 8}{u \log 3}} = \left[\lim_{u \to 0}(1+u)^{1/u}\right]^{\frac{\log 8}{\log 3}} = e^{\frac{\log 8}{\log 3}}

Since

\frac{\log 8}{\log 3} = \log_3 8

, we get

L = e^{\log_3 8}

Answer:

e^{\log_3 8}

, i.e. Option A.

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