Find x: \( 4^{x-2}+2 \times 2^{2 x-1}=1 \frac{1}{16} \)

Given: $4^{x-2}+2 \times 2^{2 x-1}=1\frac{1}{16}$

To find: $x$

Apply exponent rule: $a^{b} \times a^{c}=a^{b+c} $

$2^{2 x-1} \times 2=2^{2 x-1+1}$

$4^{x-2}+2^{2 x-1+1}=1+\frac{1}{16}$

= $2^{2(x-2)}+2^{2 x-1+1}=1+\frac{1}{16}$

Apply exponent rule: $a^{b+c}=a^{b} a^{c}$

$2^{2(x-2)}=2^{2 x} \times 2^{-4}$

$2^{2 x} \times 2^{-4}+2^{2 x-1+1}=1+\frac{1}{16}$

Apply exponent rule: $a^{b c}=\left(a^{b}\right)^{c}$

$2^{2 x}=\left(2^{x}\right)^{2}, 2^{2 x-1+1}=\left(2^{x}\right)^{2}$

$\left(2^{x}\right)^{2} \times 2^{-4}+\left(2^{x}\right)^{2}=1+\frac{1}{16}$

Rewrite the equation with $2^{x}=u$

$(u)^{2} \times 2^{-4}+(u)^{2}=1+\frac{1}{16}$

Solve $u^{2} \times 2^{-4}+u^{2}=1+\frac{1}{16}: \quad u=\sqrt{1}, u=-\sqrt{1}$

$u=\sqrt{1}, u=-\sqrt{1}$

Substitute back $u=2^{x},$ solve for $x$

Solve $2^{x}=\sqrt{1}: x=0$

$x=0$