Answer:
[tex]\huge\boxed{a^{b+2}=a^2x=x^{\frac{b+2}{b}}}[/tex]
Step-by-step explanation:
[tex]a^b=x\qquad(*)\\\\a^b=x\to\left(a^b\right)^\frac{1}{b}=x^\frac{1}{b}\\\\\text{use}\ (a^n)^m=a^{nm}\\\\a^{(b)\left(\frac{1}{b}\right)}=x^\frac{1}{b}\to a=x^\frac{1}{b}\qquad(**)\\\\a^{b+2}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\a^{b+2}=a^b\cdot a^2\\\\\text{From (*) and (**)}:\\\\a^{b+2}=x\cdot\left(x^\frac{1}{b}\right)^2\\\\\text{use}\ (a^n)^m=a^{nm}\\\\a^{b+2}=x\cdot x^{\left(\frac{1}{b}\right)(2)}\\\\a^{b+2}=x\cdot x^{\frac{2}{b}}\\\\\text{use}\ a^n\cdot a^m=a^{n+m}[/tex]
[tex]a^{b+2}=x^{1+\frac{2}{b}}\\\\a^{b+2}=x^{\frac{b}{b}+\frac{2}{b}}\\\\a^{b+2}=x^{\frac{b+2}{b}}[/tex]
