How do I solve $(1/4)^{x+1}+(1/4)^x=20$?

Question

I already know that I'm supposed to get is $5/4(1/4)^x$ after factoring, but I have absolutely no idea how to get there.

Here's what I've done thus far,

$(1/2^2)^{x+1}+(1/2^2)^x$

Not sure from there what to do; any help would be appreciated

Answer 1

\begin{eqnarray} (1/4)^{x+1}+(1/4)^x&=&20\\ (1/4)^x(1/4)^1+(1/4)^x&=&20\\ (1/4)^x(1/4+1)&=&20\\ (1/4)^x(5/4)&=&20\\ (1/4)^x&=&80/5\\ \log_2\left((1/4)^x\right)&=&\log_2(16)\\ x\log_2(1/4)&=&\log_2(16)\\ x&=&\frac{\log_2(16)}{\log_2(1/4)}\\ x&=&\frac{\log_2(2^4)}{\log_2(2^{-2})}\\ x&=&-\frac{4}{2}=-2 \end{eqnarray}

Answer 2

Isolate $\left(\frac14\right)^x \Rightarrow\left(\frac14\right)^x(\frac14+1) = \left(\frac14\right)^x\cdot\frac54 = 20$

So, $4^{-x} = 16$. Now, $x =?$

Answer 3

You have, on the left hand side, $$\left(\frac 14\right)^{x+1}\cdot\left(\frac 14\right)^x$$

$$ = \left(\frac 14\right)^x\cdot \left(\frac 14 + 1\right)= \frac 54 \left(\frac 14\right)^x = 20 \iff \left(\frac 14\right)^x = 16 \iff 4^{-x} = 16$$

Now take $\log_2$ of each side.

Answer 4

Let $y=(\frac{1}{4})^x$

$\frac{5y}{4}=20$

$y=16$

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