Question

Simplify the following logarithmic expression using logarithmic identities: $\dfrac{{{\log }_{25}}23}{{{\log }_{49}}19}-\dfrac{{{\log }_{125}}23}{{{\log }_{343}}19}$ \[\]
A.${{\log }_{7}}25$ \[\]
B.$0$ \[\]
C.${{\log }_{5}}7$ \[\]
D.${{\log }_{7}}5$ \[\]

Answer 1

Hint: We change the base for logarithms involved in the given expression to new base $d$ using the base change formula ${{\log }_{b}}x=\dfrac{{{\log }_{d}}x}{{{\log }_{d}}b}$ and then we express 25,49,125,343 as powers of 5 and 7. We use logarithmic identity of power ${{\log }_{b}}{{x}^{m}}=m{{\log }_{b}}x$ to proceed and then cancel out like terms to simplify. \[\]

Complete step-by-step answer:
We know that the logarithm is the inverse operation to 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 the condition that the argument of logarithm $x$ is always a positive number and $b$ is a positive number excluding 1. If we want to change the base of the logarithms to new base say $d>0,d\ne 1$ then we can do it using following formula,
\[{{\log }_{b}}x=\dfrac{{{\log }_{d}}x}{{{\log }_{d}}b}\]
We know from logarithmic identity of power that for some $m\ne 0$ we have,
\[{{\log }_{b}}{{x}^{m}}=m{{\log }_{b}}x\]
We are given the following expression in logarithm given in the question,
$\dfrac{{{\log }_{25}}23}{{{\log }_{49}}19}-\dfrac{{{\log }_{125}}23}{{{\log }_{343}}19}$
We change the base of all the logarithms to the base $d$. We have in the numerator of for the first term in the logarithm ${{\log }_{25}}23$ Here the original base is $b=25$ and the argument of logarithm is $x=23$. We change its base to $d$ using the base change formula to have
\[{{\log }_{25}}23=\dfrac{{{\log }_{d}}23}{{{\log }_{d}}25}=\dfrac{{{\log }_{d}}23}{{{\log }_{d}}5\times 5}=\dfrac{{{\log }_{d}}23}{{{\log }_{d}}{{5}^{2}}}\]
We use the logarithmic identity of power for $m=2$ to have
\[{{\log }_{25}}23=\dfrac{{{\log }_{d}}23}{{{\log }_{d}}25}=\dfrac{{{\log }_{d}}23}{{{\log }_{d}}5\times 5}=\dfrac{{{\log }_{d}}23}{{{\log }_{d}}{{5}^{2}}}=\dfrac{{{\log }_{d}}23}{2{{\log }_{d}}5}\]
We similarly change base to $d$ and use the logarithmic identity of power for $m=2,3,3$to have for other logarithms in the expression.
\[ \begin{align}
  & \Rightarrow {{\log }_{49}}19=\dfrac{{{\log }_{d}}19}{{{\log }_{d}}49}=\dfrac{{{\log }_{d}}19}{{{\log }_{d}}{{7}^{2}}}=\dfrac{{{\log }_{d}}19}{2{{\log }_{d}}7} \\
 & \Rightarrow {{\log }_{125}}23=\dfrac{{{\log }_{d}}23}{{{\log }_{d}}125}=\dfrac{{{\log }_{d}}23}{{{\log }_{d}}{{5}^{3}}}=\dfrac{{{\log }_{d}}23}{3{{\log }_{d}}5} \\
 & \Rightarrow {{\log }_{343}}19=\dfrac{{{\log }_{d}}19}{{{\log }_{d}}343}=\dfrac{{{\log }_{d}}19}{{{\log }_{d}}{{7}^{3}}}=\dfrac{{{\log }_{d}}19}{3{{\log }_{d}}7} \\
\end{align} \]

We put the simplified values of the logarithms in the given expression to have,
\[ \begin{align}
  &\Rightarrow \dfrac{{{\log }_{25}}23}{{{\log }_{49}}19}-\dfrac{{{\log }_{125}}23}{{{\log }_{343}}19} \\
 & \Rightarrow \dfrac{\dfrac{{{\log }_{d}}23}{2{{\log }_{d}}5}}{\dfrac{{{\log }_{d}}19}{2{{\log }_{d}}7}}-\dfrac{\dfrac{{{\log }_{d}}23}{3{{\log }_{d}}5}}{\dfrac{{{\log }_{d}}19}{3{{\log }_{d}}7}} \\
 & \Rightarrow \dfrac{{{\log }_{d}}23\times {{\log }_{d}}7}{{{\log }_{d}}19\times {{\log }_{d}}5}-\dfrac{{{\log }_{d}}23\times {{\log }_{d}}7}{{{\log }_{d}}19\times {{\log }_{d}}5} \\
 & =0 \\
\end{align}\]

So, the correct answer is “Option B”.

Note: We can alternatively solve by using base raised to a power formula ${{\log }_{{{b}^{k}}}}x=\dfrac{1}{k}{{\log }_{b}}x$ where power $k$ is a non-zero real number. We note to be careful while changing to the new base $d$ and NOT exchange the numerator and denominator. We used the prime factorization of numbers 25, 125, 49,343 to express them as powers of 5 and 7.