Question

# What is the value of ${{\log }_{0.5}}16$ is equal to ?

Hint: Apply properties of logarithm to simplify the logarithm given, which is equal to ${{\log }_{0.5}}16$. Further simplification will lead you to an expression that is written only in terms of 2, or its powers. A bit more solving will give you the answer.

The logarithm is the inverse function of exponentiation. That means the logarithm of a given number $x$ is the exponent to which another fixed number, the base $b$, must be raised, to produce that number $x$. In the simplest case, the logarithm counts the number of occurrences of the same factor in repeated multiplication; e.g., since $1000=10\times 10\times 10={{10}^{3}}$, the "logarithm base $10$" of $1000$ is $3$, or ${{\log }_{10}}(1000)=3$. The logarithm of $x$ to base $b$ is denoted as ${{\log }_{b}}x$, or without parentheses, ${{\log }_{b}}x$ .
Among all choices for the base, three are particularly common. These are $b=10,e$ (the irrational mathematical constant $\approx 2.171828$), and $b=2$ (the binary logarithm). In mathematical analysis, the logarithm base e is widespread because of the analytical properties explained below. On the other hand, base-$10$ logarithms are easy to use for manual calculations in the decimal number system.
Now in question, it is given that we have to find the value of ${{\log }_{0.5}}16$.
So now we know the property ${{\log }_{t}}y=\dfrac{{{\log }_{e}}y}{{{\log }_{e}}t}$.
So applying the property we get,
${{\log }_{0.5}}16=\dfrac{\log 16}{\log 0.5}$
Now we know the property$\log 0.2=\log (2\times {{10}^{-1}})=\log (2)+\log {{(10)}^{-1}}$,
So for$\log (0.5)=\log (5\times {{10}^{-1}})=\log 5+\log {{10}^{-1}}$.
So ${{\log }_{0.5}}16=\dfrac{\log 16}{\log 5+\log {{10}^{-1}}}$
So again simplifying in a simple manner we get,
${{\log }_{0.5}}16=\dfrac{\log 16}{\log 5+\log {{10}^{-1}}}=\dfrac{\log {{2}^{4}}}{\log 5+\log {{10}^{-1}}}$
We know the property $\log {{a}^{b}}=b\log a$.
So applying the above property we get,
$=\dfrac{\log {{2}^{4}}}{\log 5+\log {{10}^{-1}}}=\dfrac{4\log 2}{\log 5-\log 10}$
We know the property$\log a-\log b=\log \dfrac{a}{b}$.
So we get,
$=\dfrac{4\log 2}{\log 5-\log 10}=\dfrac{4\log 2}{\log \dfrac{5}{10}}=\dfrac{4\log 2}{\log \dfrac{1}{2}}$
So now again applying the above property we get,
$=\dfrac{4\log 2}{\log \dfrac{1}{2}}=\dfrac{4\log 2}{\log 1-\log 2}$
We know the value of it$\log 1=0$.
So substituting the value we get,
$=\dfrac{4\log 2}{\log 1-\log 2}=\dfrac{4\log 2}{0-\log 2}$
So simplifying in a simple manner we get,
$=\dfrac{4\log 2}{0-\log 2}=\dfrac{4\log 2}{-\log 2}=-4$
So we get the final answer as that is the value of${{\log }_{0.5}}16=-4$.

Note: Read the question carefully. Donâ€™t jumble yourself. There are many properties of logarithms we should be familiar with. Donâ€™t confuse yourself. Donâ€™t miss any term while solving, take utmost care of that.