# Find the value of b which satisfies the following equation:

${{\log }_{e}}2.{{\log }_{b}}625={{\log }_{10}}16.{{\log }_{e}}10$

A) $b=10$

B) $b=5$

C) $b=6$

D) $b=1$

Answer

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Hint: Any logarithm of the form ${{\log }_{x}}y$ can be written as $\dfrac{{{\log }_{e}}y}{{{\log }_{e}}x}$. Any logarithm of the form $\log {{a}^{n}}$ can be written as $n\log a$.

Given- ${{\log }_{e}}2.{{\log }_{b}}625={{\log }_{10}}16.{{\log }_{e}}10$

We can see that the bases of the logarithms in the question are different. First, we should make all the bases the same so that solving the question will be easier .

We know, ${{\log }_{x}}y=\dfrac{{{\log }_{e}}y}{{{\log }_{e}}x}$

So, we can write ${{\log }_{b}}625\text{ as }\dfrac{{{\log }_{e}}625}{{{\log }_{e}}b}\text{ and }{{\log }_{10}}16\text{ as }\dfrac{{{\log }_{e}}16}{{{\log }_{e}}10}$

So, ${{\log }_{e}}2.{{\log }_{b}}625={{\log }_{10}}16.{{\log }_{e}}10$ can be written as ${{\log }_{e}}\text{2}\text{.}\dfrac{{{\log }_{e}}625}{{{\log }_{e}}b}=\dfrac{{{\log }_{e}}16}{{{\log }_{e}}10}.{{\log }_{e}}10$

$\Rightarrow {{\log }_{e}}2.\dfrac{{{\log }_{e}}625}{{{\log }_{e}}b}={{\log }_{e}}16$

Now we also know that $16$ can be written as ${{2}^{4}}$ and $625$ can be written as ${{5}^{4}}$.

So, we can rewrite ${{\log }_{e}}2.\dfrac{{{\log }_{e}}625}{{{\log }_{e}}b}={{\log }_{e}}16$ as ${{\log }_{e}}2.\dfrac{{{\log }_{e}}{{5}^{4}}}{{{\log }_{e}}b}={{\log }_{e}}{{2}^{4}}$ .

Now, we know $\log {{a}^{n}}=n\log a$

So, we can write ${{\log }_{e}}{{5}^{4}}$as $4.{{\log }_{e}}5$ and ${{\log }_{e}}{{2}^{4}}$ as $4.{{\log }_{e}}2$.

Now, we have ${{\log }_{e}}2.4.\dfrac{{{\log }_{e}}5}{{{\log }_{e}}b}=4.{{\log }_{e}}2$.

We can see that we can cancel $4.{{\log }_{e}}2$ from both sides of the equation.

On cancelling $4.{{\log }_{e}}2$ from both sides of the equation, we get

$\Rightarrow \dfrac{{{\log }_{e}}5}{{{\log }_{e}}b}=1$

Now, we can take ${{\log }_{e}}b$ to the right hand side of the equation.

On taking ${{\log }_{e}}b$ to the right hand side of the equation, we get

$\Rightarrow {{\log }_{e}}5={{\log }_{e}}b$

Now, since the base of the logarithm is the same and the values of the logarithms are equal, hence, the values of the arguments of the logarithms also must be equal.

So, $b=5$.

Hence, the value of $b$ satisfying ${{\log }_{e}}2.{{\log }_{b}}625={{\log }_{10}}16.{{\log }_{e}}10$ is $b=5$.

Note: Always remember $\dfrac{\log a}{\log b}$ is not the same as $\log \left( \dfrac{a}{b} \right)$. Students generally make this mistake and get their answer wrong. $\log \left( \dfrac{a}{b} \right)$ is equal to \[\log (b)-\log (a)\] which is not the same as $\dfrac{\log a}{\log b}$. So, such mistakes should be avoided.

Given- ${{\log }_{e}}2.{{\log }_{b}}625={{\log }_{10}}16.{{\log }_{e}}10$

We can see that the bases of the logarithms in the question are different. First, we should make all the bases the same so that solving the question will be easier .

We know, ${{\log }_{x}}y=\dfrac{{{\log }_{e}}y}{{{\log }_{e}}x}$

So, we can write ${{\log }_{b}}625\text{ as }\dfrac{{{\log }_{e}}625}{{{\log }_{e}}b}\text{ and }{{\log }_{10}}16\text{ as }\dfrac{{{\log }_{e}}16}{{{\log }_{e}}10}$

So, ${{\log }_{e}}2.{{\log }_{b}}625={{\log }_{10}}16.{{\log }_{e}}10$ can be written as ${{\log }_{e}}\text{2}\text{.}\dfrac{{{\log }_{e}}625}{{{\log }_{e}}b}=\dfrac{{{\log }_{e}}16}{{{\log }_{e}}10}.{{\log }_{e}}10$

$\Rightarrow {{\log }_{e}}2.\dfrac{{{\log }_{e}}625}{{{\log }_{e}}b}={{\log }_{e}}16$

Now we also know that $16$ can be written as ${{2}^{4}}$ and $625$ can be written as ${{5}^{4}}$.

So, we can rewrite ${{\log }_{e}}2.\dfrac{{{\log }_{e}}625}{{{\log }_{e}}b}={{\log }_{e}}16$ as ${{\log }_{e}}2.\dfrac{{{\log }_{e}}{{5}^{4}}}{{{\log }_{e}}b}={{\log }_{e}}{{2}^{4}}$ .

Now, we know $\log {{a}^{n}}=n\log a$

So, we can write ${{\log }_{e}}{{5}^{4}}$as $4.{{\log }_{e}}5$ and ${{\log }_{e}}{{2}^{4}}$ as $4.{{\log }_{e}}2$.

Now, we have ${{\log }_{e}}2.4.\dfrac{{{\log }_{e}}5}{{{\log }_{e}}b}=4.{{\log }_{e}}2$.

We can see that we can cancel $4.{{\log }_{e}}2$ from both sides of the equation.

On cancelling $4.{{\log }_{e}}2$ from both sides of the equation, we get

$\Rightarrow \dfrac{{{\log }_{e}}5}{{{\log }_{e}}b}=1$

Now, we can take ${{\log }_{e}}b$ to the right hand side of the equation.

On taking ${{\log }_{e}}b$ to the right hand side of the equation, we get

$\Rightarrow {{\log }_{e}}5={{\log }_{e}}b$

Now, since the base of the logarithm is the same and the values of the logarithms are equal, hence, the values of the arguments of the logarithms also must be equal.

So, $b=5$.

Hence, the value of $b$ satisfying ${{\log }_{e}}2.{{\log }_{b}}625={{\log }_{10}}16.{{\log }_{e}}10$ is $b=5$.

Note: Always remember $\dfrac{\log a}{\log b}$ is not the same as $\log \left( \dfrac{a}{b} \right)$. Students generally make this mistake and get their answer wrong. $\log \left( \dfrac{a}{b} \right)$ is equal to \[\log (b)-\log (a)\] which is not the same as $\dfrac{\log a}{\log b}$. So, such mistakes should be avoided.

Last updated date: 20th Sep 2023

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