Question

Given $\log 2=0.3010$ and $\log 3=0.4771$ find the value of $\log \sqrt[3]{\dfrac{{{3}^{2}}{{5}^{4}}}{\sqrt{2}}}$. $$$$

Answer 1

We simplify the numerical expression first by using the logarithmic identity involving power $m{{\log }_{b}}x={{\log }_{b}}{{x}^{m}}$, then using logarithmic identity involving quotient ${{\log }_{b}}\left( \dfrac{m}{n} \right)={{\log }_{b}}m-{{\log }_{b}}n$ and using logarithmic identity involving product ${{\log }_{b}}\left( mn \right)={{\log }_{b}}m+{{\log }_{b}}n$. We simplify until all powers are removed and then we put $5=\dfrac{10}{2}$. We put the given values of $\log 2,\log 3$ to find the required value. $$$$

Complete step-by-step solution:
We know that the logarithm is the inverse operation 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 $x$, which means if ${{b}^{y}}=x$ then the logarithm denoted as log and calculated as
${{\log }_{b}}x=y$
Here $x,y,b$ are real numbers subjected to condition $x> 0$ and $b> 0,b\ne 1$
We know that
${{\log }_{b}}b=1$
We know the logarithmic identity involving power $m\ne 0$ where $m$ is real number as
$m{{\log }_{b}}x={{\log }_{b}}{{x}^{m}}$
We also know the logarithmic identity involving product as
${{\log }_{b}}\left( mn \right)={{\log }_{b}}m+{{\log }_{b}}n$
We also know the logarithmic identity involving quotient as
${{\log }_{b}}\left( \dfrac{m}{n} \right)={{\log }_{b}}m-{{\log }_{b}}n$
We are given in the question the values $\log 2=0.3010$ and $\log 3=0.4771$ to evaluate the numerical expression,
$\log \sqrt[3]{\dfrac{{{3}^{2}}{{5}^{4}}}{\sqrt{2}}}$
Here base is $b=10$. We use the logarithmic identity involving power for $x=\dfrac{{{3}^{2}}{{5}^{4}}}{\sqrt{2}},m=\dfrac{1}{3}$ to have
$\log \sqrt[3]{\dfrac{{{3}^{2}}{{5}^{4}}}{\sqrt{2}}}=\log {{\left( \dfrac{{{3}^{2}}{{5}^{4}}}{\sqrt{2}} \right)}^{\dfrac{1}{3}}}=\dfrac{1}{3}\log \left( \dfrac{{{3}^{2}}{{5}^{4}}}{\sqrt{2}} \right)$
We use logarithmic identity involving quotient in the above step for $m={{3}^{2}}{{5}^{4}},n=\sqrt{2}$ to have
$\Rightarrow \dfrac{1}{3}\left[ \log \left( {{3}^{2}}{{5}^{4}} \right)-\log \sqrt{2} \right]$
We use logarithmic identity involving product in the first term for $m={{3}^{2}},n={{5}^{4}}$ in the above step to have,
$\Rightarrow \dfrac{1}{3}\left[ \left( \log {{3}^{2}}+\log {{5}^{4}} \right)-\log {{2}^{\dfrac{1}{2}}} \right]$
We use the logarithmic identity involving power for $x=3,m=2$ in the first term, for $x=5,m=4$ in the second term and $x=2,m=\dfrac{1}{2}$ in the above step to have,
$\Rightarrow \dfrac{1}{3}\left[ \left( 2\log 3+4\log 5 \right)-\dfrac{1}{2}\log 2 \right]$
Let us replace $5=\dfrac{10}{2}$ in the above step to have,
$\Rightarrow \dfrac{1}{3}\left[ \left( 2\log 3+4\log \dfrac{10}{2} \right)-\dfrac{1}{2}\log 2 \right]$
We use logarithmic identity involving quotient in the above step for $m=10,n=2$, So we have
\Rightarrow \dfrac{1}{3}\left[ \left\\{ 2\log 3+4\left( \log 10-\log 2 \right) \right\\}-\dfrac{1}{2}\log 2 \right]
We are given the values $\log 2=0.3010$ and $\log 3=0.4771$ in the question itself. We know that for the base 10 , $\log 10={{\log }_{10}}10=1$. We put these values in the above step to have,

&\Rightarrow \dfrac{1}{3}\left[ \left\\{ 2\times 0.4771+4\left( 1-0.3010 \right) \right\\}-0.5\times 0.3010 \right] \\\ & \Rightarrow \dfrac{3.5997}{3}=1.1999 \\\ \end{aligned}$$ The above value is the required value.$$$$ **Note:** We must be careful about making the mistakes that happen while placing the values of $\log 2,\log 3$, and the identity of power $x,m$ because of the interchange. The logarithm with base 10 is called a common logarithm and similarly the logarithm with base $e$ is called natural logarithm and base 2 is called the binary logarithm. The logarithmic identity of root is ${{\log }_{b}}\sqrt[p]{x}=\dfrac{{{\log }_{b}}x}{p}$.