Equation with different bases (exponential)

I seem to be stuck with the following equation right here: $$2^x + 2^{x+1} = 3^{x+2} + 3^{x+3}$$

• It would be helpful to point out what work you have done and what has you stumped. Commented Dec 3, 2015 at 23:19
• I formed it to this: 2^x + 2^x*2^1 = 3^x*3^2 + 3^x*3^3. After that I got 2^x + 4^x = 27^x + 81^x. I used the ln-Function afterwards with which I got xln2 + xln4 = xln27 + xln81. After that I did not know further.
– user295683
Commented Dec 3, 2015 at 23:20
• @kevinzakka Regarding your edit: I don't think that's what s/he meant. Commented Dec 3, 2015 at 23:22
• Did you mean $2^x+2^{x+1}=3^{x+2}+3^{x+3}$? Commented Dec 3, 2015 at 23:23
• Yes. I meant that.
– user295683
Commented Dec 3, 2015 at 23:24

Note that $2^x+2^{x+1}=2^x+2\cdot 2^x=3\cdot 2^x$.

Similarly, $3^{x+2}+3^{x+3}=9\cdot 4\cdot 3^x$.

So our equation can be rewritten as $3\cdot 2^x=9\cdot 4\cdot 3^x$, or equivalently $(3/2)^x=1/12$, which can be solved using logarithms.

Notice, $$2^x+2^{x+1}=3^{x+2}+3^{x+3}$$ $$2^x+2\cdot 2^x=3^2\cdot 3^x+3^3\cdot 3^x$$ $$2^x+2\cdot 2^x=9\cdot 3^x+27\cdot 3^x$$ $$3\cdot 2^x=36\cdot 3^x$$ $$\frac{3^x}{2^x}=\frac{1}{12}$$

$$\left(\frac{3}{2}\right)^x=\frac{1}{12}$$ $$x\ln\frac{3}{2}=-\ln 12$$ $$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{x=-\frac{\ln 12}{\ln(3/2)}}}$$ 

The main idea is to factor each side.

Notice that $2^x + 2^{x+1} = 2^x (1 + 2^1) = 3 \cdot 2^x$.

You can do the same thing with the right-hand side. That should help a lot.

\begin{align} 2^x+2^{x+1}&=3^{x+2}+3^{x+3}\\ 2^x+2\cdot2^x&=3^2\cdot3^x+3^3\cdot3^x\\ (1+2)2^x&=(3^2+3^3)3^x\\ 3\cdot2^x&=36\cdot3^x\\ 2^x&=12\cdot3^x\\ \left(\frac23\right)^x&=12\\ \end{align} Now use logarithms.