Using exponent laws to simplify $2^t\times4^2+8^{t-2}$

Question

This is one of the questions in my maths text book and I don't quite get it, can someone please explain it and show the working out?

use index laws to find $2^t\times4^2+8^{t-2}$

Observation

The index/exponent laws are $a^xa^y=a^{x+y}$ and so on. But here the indices (exponents) $t$, $2$, $t-2$ are all on top of different bases: $2$, $4$, $8$. How to proceed?

Answer 1

$$2^t\times 4^2+8^{t-2}$$ $$2^t\times ({2^2})^2+({2^3})^{t-2}$$ $$2^t\times 2^4+2^{3t-6}$$ $$2^{t+4}+2^{3t-6}$$

From here you can find different type of expression like:$$2^t(2^4+2^{2t-6})$$