Question

Solve: ${{\log }_{e}}\left( x-a \right)+{{\log }_{e}}\left( x \right)=1$. $\left( a>0 \right)$.

Answer 1

Hint: We have been given a sum of two logarithms. Using the theorem of ${{\log }_{x}}a+{{\log }_{x}}b={{\log }_{x}}ab$, we convert it into a single logarithm. Then we use theorem ${{\log }_{x}}a=b\Rightarrow {{x}^{b}}=a$ to find the quadratic equation. We solve the equation to find the solution of the problem.

Complete step by step answer:
We have to find the value of x for the equation ${{\log }_{e}}\left( x-a \right)+{{\log }_{e}}\left( x \right)=1$.
We are going to use some logarithmic formula where ${{\log }_{x}}a+{{\log }_{x}}b={{\log }_{x}}ab$.
The base of the logarithm has to be the same.
For our given equation the left-hand side equation can be changed as
$\begin{align}
  & {{\log }_{e}}\left( x-a \right)+{{\log }_{e}}\left( x \right)=1 \\
 & \Rightarrow {{\log }_{e}}\left[ x\left( x-a \right) \right]=1 \\
\end{align}$
Also, we have ${{\log }_{x}}a=b\Rightarrow {{x}^{b}}=a$.
Now we apply the theorem on the equation ${{\log }_{e}}\left[ x\left( x-a \right) \right]=1$.
$\begin{align}
  & {{\log }_{e}}\left[ x\left( x-a \right) \right]=1 \\
 & \Rightarrow x\left( x-a \right)={{e}^{1}}=e \\
\end{align}$
Now we know $\left( a>0 \right)$.
We form the quadratic equation ${{x}^{2}}-ax-e=0$.
We have the formula of quadratic equation solving as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4pc}}{2p}$ for general equation $p{{x}^{2}}+bx+c=0$. For our equation $p=1,b=-a,c=-e$.
Using the form, we get $x=\dfrac{-\left( -a \right)\pm \sqrt{{{\left( -a \right)}^{2}}-4\times 1\times \left( -e \right)}}{2\times 1}=\dfrac{a\pm \sqrt{{{a}^{2}}+4e}}{2}$.
As $\left( a>0 \right)$, we can find that ${{a}^{2}}>0$. This means ${{a}^{2}}+4e>0$. So, the roots of the term $\sqrt{{{a}^{2}}+4e}$ won’t be complex.
But when we are taking $x=\dfrac{a-\sqrt{{{a}^{2}}+4e}}{2}$, the value of $x-a=\dfrac{a-\sqrt{{{a}^{2}}+4e}}{2}-a=\dfrac{-a-\sqrt{{{a}^{2}}+4e}}{2}<0$ becomes negative. We know logarithm can’t have value for less than or equal to 0. So, the root of $x=\dfrac{a-\sqrt{{{a}^{2}}+4e}}{2}$ is not possible.
So, the roots $x=\dfrac{a+\sqrt{{{a}^{2}}+4e}}{2}$ of the equation ${{x}^{2}}-ax-e=0$ is the solution of the function ${{\log }_{e}}\left( x-a \right)+{{\log }_{e}}\left( x \right)=1$.

Note: Both a and e are constant. We don’t need to break $x=\dfrac{a+\sqrt{{{a}^{2}}+4e}}{2}$ to find the exact answer. Also, we need to always put the value of the root to double check the answer of the equation. As $\sqrt{{{a}^{2}}+4e}>a$, that’s why the value of $x-a=\dfrac{a+\sqrt{{{a}^{2}}+4e}}{2}-a=\dfrac{\sqrt{{{a}^{2}}+4e}-a}{2}>0$ is positive.