Answer::
[tex]x=\frac{2}{\sqrt{e-1}}\text{ or }x=-\frac{2}{\sqrt[]{e-1}}[/tex]Explanation:
Given the equation:
[tex]\ln (x^2+4)-2\ln (x)=1[/tex]First, we can make the number 2 an index:
[tex]ln(x^2+4)-\ln (x^2)=1[/tex]Apply the division law of logarithm to combine the left-hand side:
[tex]\ln (\frac{x^2+4}{x^2})=1[/tex]Take the exponent of both sides:
[tex]\begin{gathered} e^{\ln (\frac{x^2+4}{x^2})}=e^1 \\ \frac{x^2+4}{x^2}=e \\ x^2+4=ex^2 \end{gathered}[/tex]Solve the equation above for x:
[tex]\begin{gathered} ex^2-x^2=4 \\ x^2(e-1)=4 \\ x^2=\frac{4}{e-1} \\ x=\pm\sqrt[]{\frac{4}{e-1}} \\ x=\pm\frac{2}{\sqrt[]{e-1}} \end{gathered}[/tex]The values of the variable x are:
[tex]x=\frac{2}{\sqrt{e-1}}\text{ or }x=-\frac{2}{\sqrt[]{e-1}}[/tex]