# How to Solve a Quadratic Logarithmic Equation?

# Can you solve the following quadratic logarithmic equation?

If the square of the natural logarithm of “1 plus 4 by *x*” plus the square of the natural logarithm of “1 minus 4 by (*x* + 4)” is equal to 2 times the square of the natural logarithm of “2 by (*x*-1)minus 1″, then solve this find the value of *x * quadratic logarithmic equation or find the value of* x*

\bigg(\log\bigg(1+\dfrac{4}{x}\bigg)\bigg)^2+\bigg(\log\bigg(1-\dfrac{4}{x+4}\bigg)\bigg)^2=2\bigg(\log\bigg(\dfrac{2}{x-1}-1\bigg)\bigg)^2

x=??

## Solution: Solving the logarithmic equation

Given equation is

\begin{aligned} &\bigg(\log\bigg(1+\dfrac{4}{x}\bigg)\bigg)^2+\bigg(\log\bigg(1-\dfrac{4}{x+4}\bigg)\bigg)^2=2\bigg(\log\bigg(\dfrac{2}{x-1}-1\bigg)\bigg)^2\\ \\ \Rightarrow \ \ & \bigg(\log\bigg(\dfrac{x+4}{x}\bigg)\bigg)^2+\bigg(\log\bigg(\dfrac{x+4-4}{x+4}\bigg)\bigg)^2=2\bigg(\log\bigg(\dfrac{2-x+1}{x-1}\bigg)\bigg)^2\\ \\ \Rightarrow \ \ & \bigg(\log\bigg(\dfrac{x+4}{x}\bigg)\bigg)^2+\bigg(\log\bigg(\dfrac{x}{x+4}\bigg)\bigg)^2=2\bigg(\log\bigg(\dfrac{3-x}{x-1}\bigg)\bigg)^2\\ \\ \end{aligned}

We know

\begin{aligned} &\bigg(\log\bigg(\dfrac{x+4}{x}\bigg)+\log\bigg(\dfrac{x}{x+4}\bigg)\bigg)^2\\ \\ &=\bigg(\log\bigg(\dfrac{x+4}{x}\bigg)\bigg)^2+2\log \bigg(\dfrac{x+4}{x}\bigg)\log \bigg(\dfrac{x}{x+4}\bigg)+\bigg(\log\bigg(\dfrac{x}{x+4}\bigg)\bigg)^2\\ \end{aligned}

Now

\begin{aligned} &\bigg(\log\bigg(\dfrac{x+4}{x}\bigg)+\log\bigg(\dfrac{x}{x+4}\bigg)\bigg)^2\\ \\ &=\bigg(\log\bigg(\dfrac{x+4}{x}\bigg)\bigg)^2+2\log \bigg(\dfrac{x+4}{x}\bigg)\log \bigg(\dfrac{x}{x+4}\bigg)+\bigg(\log\bigg(\dfrac{x}{x+4}\bigg)\bigg)^2\\ \\ &\bigg(\log\bigg(\dfrac{x+4}{x}\bigg)+\log\bigg(\dfrac{x}{x+4}\bigg)\bigg)^2=\bigg(\log\bigg(\dfrac{x+4}{x}\bigg)-\log\bigg(\dfrac{x+4}{x}\bigg)\bigg)^2\\ \\ \Rightarrow \ \ &\bigg(\log\bigg(\dfrac{x+4}{x}\bigg)+\log\bigg(\dfrac{x}{x+4}\bigg)\bigg)^2=\bigg(\log\bigg(\dfrac{x+4}{x}\bigg)-\log\bigg(\dfrac{x+4}{x}\bigg)\bigg)^2\\ \\ \Rightarrow \ \ &\bigg(\log\bigg(\dfrac{x+4}{x}\bigg)+\log\bigg(\dfrac{x}{x+4}\bigg)\bigg)^2=0\\ \end{aligned}

So

\begin{aligned} &0=\bigg(\log\bigg(\dfrac{x+4}{x}\bigg)\bigg)^2+2\log \bigg(\dfrac{x+4}{x}\bigg)\log \bigg(\dfrac{x}{x+4}\bigg)+\bigg(\log\bigg(\dfrac{x}{x+4}\bigg)\bigg)^2\\ \\ \Rightarrow \ \ & \bigg(\log\bigg(\dfrac{x+4}{x}\bigg)\bigg)^2+\bigg(\log\bigg(\dfrac{x}{x+4}\bigg)\bigg)^2 = -2\log \bigg(\dfrac{x+4}{x}\bigg)\log \bigg( \dfrac{x}{x+4}\bigg)\\ \end{aligned}

Then

\begin{aligned} & \bigg(\log\bigg(\dfrac{x+4}{x}\bigg)\bigg)^2+\bigg(\log\bigg(\dfrac{x}{x+4}\bigg)\bigg)^2=2\bigg(\log\bigg(\dfrac{3-x}{x-1}\bigg)\bigg)^2\\ \\ \Rightarrow \ \ & -2\log \bigg(\dfrac{x+4}{x}\bigg)\log \bigg( \dfrac{x}{x+4}\bigg)=2\bigg(\log\bigg(\dfrac{3-x}{x-1}\bigg)\bigg)^2\\ \\ \Rightarrow \ \ & -\log \bigg(\dfrac{x+4}{x}\bigg)\log \bigg( \dfrac{x}{x+4}\bigg)=\bigg(\log\bigg(\dfrac{3-x}{x-1}\bigg)\bigg)^2\\ \\ \Rightarrow \ \ & \bigg(\log \bigg(\dfrac{x+4}{x}\bigg)\bigg)^2=\bigg(\log\bigg(\dfrac{3-x}{x-1}\bigg)\bigg)^2\\ \\ \Rightarrow \ \ & \dfrac{x}{x+4}=\dfrac{3-x}{x-1}\\ \\ \Rightarrow \ \ & \dfrac{x}{x+4}=\dfrac{3-x}{x-1}\\ \\ \Rightarrow \ \ & {x}{(x-1)}={(3-x)}{(x+4)}\\ \\ \Rightarrow \ \ & x^2-x=3x+12-x^2-4x\\ \\ \Rightarrow \ \ & 2x^2= 12\\ \\ \Rightarrow \ \ & x^2= 6\\ \\ \Rightarrow \ \ & x= \pm \sqrt6\\ \\ \end{aligned}

That is, the solution of the equation is *x* = ±√6