Question

Assuming that $\log \left( mn \right)=\log m+\log n$. Prove that $\log {{x}^{n}}=n\log x$.

Answer 1

Hint: Extend the given formula for $n$number of times to get the $n$in the given problem and put $m=n=x$.

Here we are given that $\log \left( mn \right)=\log m+\log n$.
We have to prove that $\log {{x}^{n}}=n\log x$.
Now, we take the equation given.
$\log \left( mn \right)=\log m+\log n\to \text{equation}\left( i \right)$
As we can see that, we have to prove the given equation in terms of$x$.
Therefore, we put $m=n=x$in equation$\left( i \right)$.
We get, $\log \left( x.x \right)=\log x+\log x$
$=\log {{x}^{2}}=2\log x$
Now we will add $\log x$on both sides,
$=\log {{x}^{2}}+\log x=2\log x+\log x$
$=\log {{x}^{2}}.x=3\log x$[From equation$\left( i \right)$]
As we know that ${{a}^{m}}.{{a}^{n}}={{a}^{m+n}}$
Therefore, ${{x}^{2}}.{{x}^{2}}={{x}^{2+1}}={{x}^{3}}$
Hence, we get $\log {{x}^{3}}=3\log x$
Similarly, if we add \log x$$$$ntimes
We get, $\log x+\log x+\log x....n\text{ times}=\log \left( x.x.x.x....n\text{ times} \right)$[From
equation$\left( i \right)$]
We get, $n\log x=\log \left( x.x.x....n\text{ times} \right)$
As, ${{a}^{{{m}_{1}}}}.{{a}^{{{m}_{2}}}}.....{{a}^{{{m}_{n}}}}={{a}^{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}+.....{{m}_{n}}} }$
We get, ${{x}^{1}}.{{x}^{1}}.{{x}^{1}}.....n\text{ times = }{{\text{x}}^{1+1+1.....n\text{ times}}}={{x}^{n}}$
Hence, we get $n\log x=\log {{x}^{n}}$
Therefore, we proved the desired equation.

Note: Students must note that they have to prove $\log {{x}^{n}}=n\log x$starting from $\log m+\log n=\log mn$because the result can also be proved by rules of logarithm. That is, by taking ${{\log }_{a}}x=t$and putting $x={{a}^{t}}$and then raising both sides to the power of $n$which would be
wrong for a given question.