Question

Find the value of
\[\begin{align}
  & \left( i \right){{\log }_{\frac{1}{2}}}8 \\
 & \left( ii \right){{\log }_{5}}0.008 \\
 & \left( iii \right){{\log }_{5}}8 \\
 & \left( iv \right){{\log }_{7}}\sqrt[3]{7} \\
\end{align}\]

Answer 1

Hint: In this question, we need to find the value of various logarithmic functions. We will first put the function equal to some variable and then find the value of the variable using various logarithmic properties. We will use following properties:
$\left( i \right){{\log }_{b}}\left( x \right)=y\Rightarrow {{b}^{y}}=x$ (Definition of logarithm)
$\left( ii \right)\dfrac{1}{x}={{x}^{-1}}$ (Properties of exponents)
$\left( iii \right){{\log }_{a}}x=\dfrac{{{\log }_{b}}x}{{{\log }_{b}}a}$ (Change of base property)
$\begin{align}
  & \left( iv \right)\log 8=0.9030,\log 5=0.6989 \\
 & \left( v \right){{\log }_{b}}{{b}^{x}}=x \\
\end{align}$

Complete step by step answer:
Here we need to find the value of various logarithmic functions. Let us solve them one by one.
(i) ${{\log }_{\dfrac{1}{2}}}8$.
Here base is $\dfrac{1}{2}$ and power is 8. Let us suppose that ${{\log }_{\dfrac{1}{2}}}8=p$.
${{\log }_{b}}\left( x \right)=y\Rightarrow {{b}^{y}}=x$.
So here $b=\dfrac{1}{2},x=8,y=p$.
So we get ${{\log }_{\dfrac{1}{2}}}8=p\Rightarrow {{\left( \dfrac{1}{2} \right)}^{p}}=8\Rightarrow \dfrac{1}{{{2}^{p}}}=8$.
We know that $\dfrac{1}{x}={{x}^{-1}}$ so we get, ${{2}^{-p}}=8$.
Now we can write 8 as $2\times 2\times 2={{2}^{3}}$ so we get ${{2}^{-p}}={{2}^{3}}$.
By law of indices, if the base are equal than the exponents are also equal. Hence, $-p=3\Rightarrow p=-3$.
Therefore ${{\log }_{\dfrac{1}{2}}}8=-3$.
(ii) ${{\log }_{5}}0.008$.
Let us suppose that ${{\log }_{5}}0.008=p$.
We know ${{\log }_{b}}\left( x \right)=y\Rightarrow {{b}^{y}}=x$.
Here b = 5, x = 0.008, y = p, so we get,
${{\log }_{5}}0.008=p\Rightarrow {{5}^{p}}=0.008\Rightarrow {{5}^{p}}=\dfrac{8}{1000}\Rightarrow {{5}^{p}}=\dfrac{1}{125}$.
We know $\dfrac{1}{x}={{x}^{-1}}$ so we get ${{5}^{p}}={{\left( 125 \right)}^{-1}}$.
We can write 125 as $5\times 5\times 5={{5}^{3}}$ so we get, ${{5}^{p}}={{\left( {{5}^{3}} \right)}^{-1}}\Rightarrow {{5}^{p}}={{5}^{-3}}$.
By law of indices $p=-3$.
Therefore, ${{\log }_{5}}0.008=-3$.
(iii) ${{\log }_{5}}8$.
We know that, ${{\log }_{a}}x=\dfrac{{{\log }_{b}}x}{{{\log }_{b}}a}$.
Here a = 5, x = 8 taking b as 10 we get, \[{{\log }_{5}}8=\dfrac{{{\log }_{10}}5}{{{\log }_{10}}8}\].
We know ${{\log }_{10}}x$ can be written as logx so we get \[{{\log }_{5}}8=\dfrac{\log 5}{\log 8}\].
The value of log5 is 0.9030 and value of log8 is 0.6989 so we get \[{{\log }_{5}}8=\dfrac{0.9030}{0.6989}=1.2920\].
Therefore, \[{{\log }_{5}}8=1.2920\].
(iv) ${{\log }_{7}}\sqrt[3]{7}$.
We know that $\sqrt[3]{7}={{\left( 7 \right)}^{\dfrac{1}{3}}}$ so we get ${{\log }_{7}}{{\left( 7 \right)}^{\dfrac{1}{3}}}$.
We know that ${{\log }_{b}}{{b}^{x}}=x$ hence we get ${{\log }_{7}}\sqrt[3]{7}=\dfrac{1}{3}$.

Note: Students should know all the logarithmic properties before solving this sum. Students often get confused with the base and power of logarithmic. If we are given log x then it means that logarithmic has been 10. If we are given ln x then it means that it is natural log (log with base e).