Question

# If ${\log _4}\left( {{x^2} + x} \right) - {\log _4}\left( {x + 1} \right) = 2$, then the value of x is equal toA. 1B. 2C. 4D. 16

Hint: We can simplify the equation using the properties of logarithms as $\log a - \log b = \log \left( {\dfrac{a}{b}} \right)$ and ${\log _a}x = {a^x}$. Then we can simplify and take antilogarithm on both sides. Then we can solve for x to get the required solution.

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.

Note: The concept of the logarithm is used to solve this problem. By properties of logarithms,$\log a + \log b = \log \left( {ab} \right)$ and $\log a - \log b = \log \left( {\dfrac{a}{b}} \right)$. These properties of logarithm are used to do complex multiplication and division. For that, we take the logarithm of the expression, then do the operations and then we take the antilog to get the required answer. Logarithm to the base a can be defined as the power of a when the number is written as an exponent of a. Antilogarithm or exponent is the inverse operation of the logarithm.