The correct option is A 34 Given, log_3^43^3 We have log_a^na^m=mnlog_aa =34log_33 =34× 1 since log_aa=1 =34 ...

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Chain Rule of Differentiation

Solve: log8...

Question

Solve: ${log}_{81} 27$

\frac{3}{4}

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\frac{4}{3}

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\frac{27}{81}

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\frac{81}{27}

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Solution

The correct option is A $\frac{3}{4}$
Given, ${log}_{3^{4}} 3^{3}$

We have ${log}_{a^{n}} a^{m} = \frac{m}{n} {log}_{a} a$

$= \frac{3}{4} {log}_{3} 3$

$= \frac{3}{4} \times 1$ since ${log}_{a} a = 1$

$= \frac{3}{4}$

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Similar questions

Solve:

[[5×{(−3)×(−4)}]÷{27 + (−14−3)}][[3×{(−4)×(−5)}]÷{27+(−13−4)}]
+[{27×(−2)}÷(−3−3)] + [{27×(−2)}÷(−3−3)]
+[{(−3)×(−3)×5}÷(−3)]]

\sqrt[4]{27} \div \sqrt[4]{3}

{\sqrt[3]{\frac{1}{(27)^{- 4}}}}^{- \frac{3}{2}}

\frac{9^{n} \times 3^{5} \times (27)^{3}}{3 \times (81)^{4}} = 27

n

(81)^{\frac{3}{4}} = 27

{log}_{81} 27 = \frac{3}{m}

m

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