A classical algebra problem

$3^{2x+2} + 27^{x+1} = 36$

Step 1: Simplify

$3^{2x+2}$ can be simplified into $3^{2(x+1)}$ , which is equal to $(3^{2})^{x+1} = 9^{x+1}$

So now we have $9^{x+1} + 27^{x+1} = 36$

Step 2: Decide which exponents make the equation true

We already know that $9 + 27 = 36$ , so in order to make this equation true, we need for the exponents to be equal to 1 .

$\Rightarrow$ in other words, we want $9^{1} + 27^{1} = 36$ .

For the exponents to be equal to 1, we want: $x+1=1$

Step 3: Solve

The value of x that makes the above statement true is $x=0$ , so x must be zero.