Rewrite $2^{x-6} + 2^x$ as $A \cdot 2^x$

Question

I simply don't understand how to rewrite $$2^{x-6} + 2^x =A\cdot 2^x$$ for some $A.$

I can work up to a certain point but after that, I'm lost. Here is my work:

$$2^{x-6}+2^x$$ $$\to 2^x \cdot 2^{-6} + 2^x$$ $$\to 2^x \cdot \frac1{64} + 2^x$$

After that, I don't know what to do. Could someone please help me understand how to solve this?

Answer 1

You are almost there! Now all we have to do is use the distributive rule, which gives $$2^x*1/64+2^x = 2^x*(1+1/64) = 65/64 * 2^x$$ Therefore, $A = 65/64$

Answer 2

The task is to factor out $2^x$. Note that $2^{x-6} = 2^{-6} \cdot 2^x $, and similarly $2^x = 1\cdot 2^x $, therefore

$$2^{x-6} + 2^x = (2^{-6} +1) \cdot 2^x = A \cdot 2^x$$

for $A = 2^{-6} + 1$.