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 \[\log x\]\[n\]times
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.