Answer
The value of ${{\log }_{2}}x+{{\log }_{2}}\left( 2x-3 \right)=1$ is $ x=2$.
Work Step by Step
We have to find the value of x from ${{\log }_{2}}x+{{\log }_{2}}\left( 2x-3 \right)=1$ by evaluating the left-hand side as follows:
$\begin{align}
& {{\log }_{2}}x+{{\log }_{2}}\left( 2x-3 \right)=1 \\
& {{\log }_{2}}x\left( 2x-3 \right)=1 \\
& {{\log }_{2}}\left[ 2{{x}^{2}}-3x \right]=1
\end{align}$
Then, using if ${{p}^{q}}=r $, then ${{\log }_{p}}r=q $, one gets:
$\begin{align}
& 2{{x}^{2}}-3x={{2}^{1}} \\
& 2{{x}^{2}}-3x-2=0 \\
& 2{{x}^{2}}-4x+x-2=0
\end{align}$
Simplify it further:
$\begin{align}
& 2x\left( x-2 \right)+x-2=0 \\
& \left( 2x+1 \right)\left( x-2 \right)=0
\end{align}$
So, $ x=2$ or $ x=-\frac{1}{2}$.
If $ x=-\frac{1}{2}$, then ${{\log }_{2}}x $ does not exist as the logarithm function does not take on negative values.
Therefore, the only possible value of x is 2.
Thus, the value is $ x=2$.
