Question

Solve the following equations.
$\sqrt{7^{2x^2-5x-6}}=(\sqrt{2}{)}^{3lo{g}_{2}49}$

Answer 1

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

from log property

$a^{log{b}^c}=c^{log}{b}^a$ and $alo{g}_{b}c=lo{g}_b{c}^a$

we can write given Expression Like this

$\sqrt{7^{2x^2-5x-6}}=\left(\sqrt{2}\right)lo{g}_2{49}^3$ { $alo{g}_{b}c=lo{g}_b{c}^a$

$\sqrt{7^{2x^2-5x-6}}=({49}^3{)}^{lo{g}_2\sqrt{2}}$ $[a^{lo{g}_b^c}=c^{lo{g}_b^a}]$

$\sqrt{7^{2x^2-5x-6}}=({49}^3{)}^{\frac{1}{2}}lo{g}_2^2$

$\sqrt{7^{2x^2-5x-6}}=({49}^3{)}^{1/2}={49}^{3/2}=7^3=(\sqrt{7^6})$

By comparing both sides

$7^{2x^2-5x-6}=7^6$

$2x^2-5x-6=6$

$2x^2-5x-12=0\Rightarrow$ By using any find we van find roots of this Eq${}^2$.

So, $x=\frac{5\pm \sqrt{25+96}}{4}=\frac{5\pm 11}{4}$

$x=4,-3/2$

If we put both the values in initially

step then it satisfies the given $E{q}^n$

So both values are its solution.

$x=4andx=-3/2$