Question

Find the value of $x$ in ${{\log }_{e}}2\cdot {{\log }_{x}}625={{\log }_{10}}16\cdot {{\log }_{e}}10$.

Answer 1

Hint: We use the base change formula ${{\log }_{k}}x={{\log }_{b}}x\times {{\log }_{k}}b$ in the right hand side of the give expression to eliminate the base 10. We simplify the logarithm in both sides using the identity $m{{\log }_{b}}x={{\log }_{b}}{{x}^{m}}$ and the divide the common term occurring in both sides.

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 condition $x > 0$ and $b > 0,b\ne 1$
We know that
\[{{\log }_{b}}b=1\]
We know the logarithmic identity involving power $m\ne 0$ where $m$ is real number as
\[m{{\log }_{b}}x={{\log }_{b}}{{x}^{m}}\]
We also know the logarithmic identity involving quotient as
\[{{\log }_{b}}\left( \dfrac{m}{n} \right)={{\log }_{b}}m-{{\log }_{b}}n\]
We can change to base our choice $k>0,k\ne 1$ using the identity
\[\begin{align}
  & {{\log }_{b}}x=\dfrac{{{\log }_{k}}x}{{{\log }_{k}}b} \\
 & \Rightarrow {{\log }_{k}}x={{\log }_{b}}x\times {{\log }_{k}}b....\left( 1 \right) \\
\end{align}\]

When the base of the logarithm $b=e$ an transcendental irrational with approximate value $e\simeq 2.718$then the logarithm is called natural logarithm and if $b=10$ then logarithm is called common logarithm.
The given expression is
\[{{\log }_{e}}2\cdot {{\log }_{x}}625={{\log }_{10}}16\cdot {{\log }_{e}}10\]
We see that on the right hand side of the equation has two bases $b=e,10$. We use the base change identity (1) for $x=16,b=10,k=e$ to have
\[\Rightarrow {{\log }_{e}}2\cdot {{\log }_{x}}625={{\log }_{e}}16\]
We find the prime factorization of 16 in the above step and replace 16 as $16=2\times 2\times 2\times 2={{\left( 2 \right)}^{4}}$ in the above step. We similarly find prime factorization of 625 and replace 625 as $525=5\times 5\times 5\times 5={{\left( 5 \right)}^{4}}$ in the above step. We have
\[\Rightarrow {{\log }_{e}}2\cdot {{\log }_{x}}{{5}^{4}}={{\log }_{e}}{{2}^{4}}\]
We use the logarithmic identity of power $m{{\log }_{b}}x={{\log }_{b}}{{x}^{m}}$ for $x=5,b=e,m=4$ in left hand side and for $x=2,b=e,m=4$ in the right hand side of the above step to have
\[\begin{align}
  & \Rightarrow {{\log }_{e}}2\cdot \left( 4{{\log }_{x}}5 \right)=4{{\log }_{e}}2 \\
 & \Rightarrow 4{{\log }_{e}}2\cdot {{\log }_{x}}5=4{{\log }_{e}}2 \\
\end{align}\]
We divide both side of the equation by $4{{\log }_{e}}2$ because ${{\log }_{e}}2\ne 0$ and have,
\[\Rightarrow {{\log }_{x}}5=1\]
 We use the definition of logarithm and have,
\[\begin{align}
  & \Rightarrow 5={{x}^{1}} \\
 & \Rightarrow x=5 \\
\end{align}\]

Note: The base of natural logarithm can be defined as a real number $a$ such that $\log a=\int\limits_{0}^{1}{\dfrac{1}{x}}dx$ and can also be approximated by putting $x=1$in the series$1+\dfrac{x}{2}+\dfrac{{{x}^{2}}}{2!}+...$ The natural logarithm also commonly denoted as $\ln x$ is primarily used in radio decay, information theory etc.