Questions & Answers

Question

Answers

A. 1

B. 2

C. 4

D. 16

Answer
Verified

We have the equation, ${\log _4}\left( {{x^2} + x} \right) - {\log _4}\left( {x + 1} \right) = 2$

We know that $\log a - \log b = \log \left( {\dfrac{a}{b}} \right)$,

\[ \Rightarrow {\log _4}\left( {\dfrac{{{x^2} + x}}{{x + 1}}} \right) = 2\]

We know that the antilogarithm of ${\log _a}x$ is ${a^x}$. So, we can take the antilogarithm on both sides of the equation, we get,

\[ \Rightarrow {4^{{{\log }_4}\left( {\dfrac{{{x^2} + x}}{{x + 1}}} \right)}} = {4^2}\]

We know that ${a^{{{\log }_a}x}} = x$.

\[ \Rightarrow \left( {\dfrac{{{x^2} + x}}{{x + 1}}} \right) = {4^2}\]

Factoring the numerator of the LHS of the equation, we get,

\[ \Rightarrow \left( {\dfrac{{x\left( {x + 1} \right)}}{{x + 1}}} \right) = 16\]

Cancelling the common terms, we get,

$x = 16$

Therefore, the value of x is 16.

So, the correct answer is option D.

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