Question

Solve
$\sqrt{7^{2x^2}-5x-6}=(\sqrt{2}{)}^{3{\log }_{2}49}$

Answer 1

$\begin{array}{c}\sqrt{7^{2x^2}-5x-6}={\left(\sqrt{2}\right)}^{3{{\log }_2}^{49}}\\ taking\log bothsides\\ \frac{1}{2}\log \left(7^{2x^2}-5x-6\right)=\log {\left(2\right)}^{3/2}{\log }_2^{7^2}\\ \frac{1}{2}\frac{{\log }_2{7}^{2x^2}}{{\log }_2\left(5x+6\right)}=\frac{3}{2}\times 2{\log }_2^7{\log }_2^2\\ \frac{1}{2}\times \frac{2x^2{\log }_2^7}{{\log }_{1}5x+{\log }_2^6}=3{\log }_2{\log }_2^2\\ {\log }_2\left(7^{2x^2}-5x-6\right)={\log }_2^{7^6}\\ 7^{2x^2}-5x-6=7^6\\ 7^{2x^2}=7^6+6+5x\\ 2x^2=8\\ x=2\end{array}$