Question

The value of ${\log _{0.5}}16$ is equal to
A. $- 4$
B. 4
C. $\dfrac{1}{4}$
D. $- \dfrac{1}{4}$

Answer 1

Hint: As, the base of the log in the given expression is 0.5 which is equals to $\dfrac{1}{2}$ , convert 16 into powers of 2 and then write it as power of $\dfrac{1}{2}$. Then, use the formula, $\log {a^m} = m\log a$ to find the value of the given expression.

Complete step by step answer:
First of all we will convert 16 in powers of 2.
As, $16 = 2 \times 2 \times 2 \times 2$, so, 16 can be written as $16 = {2^4}$.
The number 0.5 in the expression, ${\log _{0.5}}16$ is the base of the log.
We know that, $0.5 = \dfrac{1}{2}$
Thus, the expression can be rewritten as ${\log _{\dfrac{1}{2}}}\left( {{2^4}} \right)$
Also, ${2^4} = \dfrac{1}{{{2^{ - 4}}}}$ because ${a^n} = \dfrac{1}{{{a^{ - n}}}}$
Therefore, the expression is rewritten as, ${\log _{\dfrac{1}{2}}}{\left( {\dfrac{1}{2}} \right)^{ - 4}}$
We know that ${a^m} = m\log a$.
Hence, ${\log _{\dfrac{1}{2}}}{\left( {\dfrac{1}{2}} \right)^{ - 4}}$ can be written as, $- 4{\log _{\dfrac{1}{2}}}\left( {\dfrac{1}{2}} \right)$.
We know that, ${\log _a}a = 1$, therefore, ${\log _{\dfrac{1}{2}}}\left( {\dfrac{1}{2}} \right) = 1$
On substituting the value ${\log _{\dfrac{1}{2}}}\left( {\dfrac{1}{2}} \right) = 1$ in the expression $- 4{\log _{\dfrac{1}{2}}}\left( {\dfrac{1}{2}} \right)$ , we get,
$- 4{\log _{\dfrac{1}{2}}}\left( {\dfrac{1}{2}} \right) = - 4\left( 1 \right) = - 4$
Hence, the value ${\log _{0.5}}16$ = $- 4$
Therefore, option A is correct.

Note: This question can alternatively be done by converting the logarithmic problem to exponential form. Let the given expression ${\log _{0.5}}16$ be equals to $x$. Then, the given expression can be converted as, ${\left( {0.5} \right)^x} = 16$. Solve it by writing 16 as the power of 0.5 and then comparing the values.
${\left( {0.5} \right)^x} = 16 \\\ \Rightarrow {\left( {0.5} \right)^x} = {2^4} \\\ \Rightarrow {\left( {0.5} \right)^x} = {\left( {\dfrac{1}{2}} \right)^{ - 4}} \\\ \Rightarrow{\left( {0.5} \right)^x} = {\left( {0.5} \right)^{ - 4}} \\\ \Rightarrow x = - 4 \\\$