Question

How do you solve the logarithmic equation: $3{{\log }_{5}}x-{{\log }_{5}}\left( 5x \right)=3-{{\log }_{5}}25$?

Answer 1

We solve the given equation using the different identity formulas of logarithm like $\log a+\log b=\log \left( ab \right)$, ${{\log }_{m}}a=y\Rightarrow a={{m}^{y}}$. We first take all the logarithmic functions on one side. The main step would be to eliminate the logarithm function and keep only the quadratic equation of x. we solve the equation with the help of factorisation.

Complete step by step solution:
We first take all the logarithmic functions on one side for $3{{\log }_{5}}x-{{\log }_{5}}\left( 5x \right)=3-{{\log }_{5}}25$.
So, $3{{\log }_{5}}x-{{\log }_{5}}\left( 5x \right)+{{\log }_{5}}25=3$. Using the identity $p{{\log }_{x}}a={{\log }_{x}}{{a}^{p}}$ for $3{{\log }_{5}}x$, we get
$3{{\log }_{5}}x={{\log }_{5}}{{x}^{3}}$.
We use the identity functions of $\log a+\log b=\log \left( ab \right),\log a-\log b=\log \left( \dfrac{a}{b} \right)$.
We operate the addition and subtraction part in the left-hand side of ${{\log }_{5}}{{x}^{3}}-{{\log }_{5}}\left( 5x \right)+{{\log }_{5}}25=3$. All of the logarithms have the same base.
${{\log }_{5}}{{x}^{3}}-{{\log }_{5}}\left( 5x \right)+{{\log }_{5}}25={{\log }_{5}}\left( \dfrac{{{x}^{3}}\times 25}{5x} \right)={{\log }_{5}}\left( 5{{x}^{2}} \right)$.
The equation becomes ${{\log }_{5}}\left( 5{{x}^{2}} \right)=3$.
Now we have to eliminate the logarithm function to find the quadratic equation of x.
We know ${{\log }_{m}}a=y\Rightarrow a={{m}^{y}}$.
Applying the rule in case of ${{\log }_{5}}\left( 5{{x}^{2}} \right)=3$, we get
$\begin{aligned} & {{\log }_{5}}\left( 5{{x}^{2}} \right)=3 \\\ & \Rightarrow 5{{x}^{2}}={{5}^{3}} \\\ & \Rightarrow {{x}^{2}}={{5}^{2}} \\\ & \Rightarrow x=\pm 5 \\\ \end{aligned}$
Now if we put $x=-5$, the logarithmic values become negative which is not possible.
Therefore, solution of $3{{\log }_{5}}x-{{\log }_{5}}\left( 5x \right)=3-{{\log }_{5}}25$ is $x=5$.

Note: In case of the base is not mentioned then the general solution for the base for logarithm is 10. But the base of $e$ is fixed for $\ln$.
We also need to remember that for logarithm function there has to be a domain constraint.
For any ${{\log }_{b}}a$, $a>0$. This means for ${{\log }_{5}}x,{{\log }_{5}}\left( 5x \right)$, $x>0$ and $5x>0$.
The simplified form is $x>0$.