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

Answer

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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.

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.

Last updated date: 25th Sep 2023

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