Precalculus (6th Edition) Blitzer

$x=5$
RECALL: (i) If $a^M=a^N$, then $M=N.$ (ii) $\sqrt[n]{a}=a^{\frac{1}{n}}.$ Use rule (ii) above to obtain $7^{\frac{x-2}{6}}=7^{\frac{1}{2}}.$ Use rule (i) above to obtain $\dfrac{x-2}{6}=\dfrac{1}{2}.$ Cross-multiply to obtain $\begin{array}{ccc} &2(x-2) &= &6(1) \\&2x-4 &= &6 \\&2x &= &6+4 \\&2x&= &10 \\&x &= &\dfrac{10}{2} \\&x &= &5 \\\end{array}.$