Question

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

Answer 1

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. Don’t jumble yourself. There are many properties of logarithms we should be familiar with. Don’t confuse yourself. Don’t miss any term while solving, take utmost care of that.