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

Similar Posts