# How to solve $32^x - 8 = 2 \cdot 4^x$

I am given the following equation to solve

$$32^x - 8 = 2 \cdot 4^x$$

which one can simplyfy to $$2^{5x}-2^3 = 2^{2x+1}$$

where do we go from here? If we had something like $$2^{2x} - 5 \cdot 2^x + 6 = 0$$

we could convert it to a quadratic, but not in this case.

Any help is highly appreciated.

• This leads to a difficult polynomial equation. Are you sure about the coefficients ? – Yves Daoust Oct 5 '18 at 12:46
• @YvesDaoust yes, exactly my conclusion. Yes I am sure. – bru1987 Oct 5 '18 at 13:01
• Where is that coming from ? – Yves Daoust Oct 5 '18 at 13:06
• From a colleague's material. He had that on a 10th grade, no calc, test. – bru1987 Oct 5 '18 at 13:07
• I'm thinking so too, the only real solution isn't a rational root as found by wolfram. – Diehardwalnut Oct 5 '18 at 13:27

You can still convert it into a polynomial, since

$$2^{5x} - 8 = 2\cdot 2^{2x}$$

converts to

$$y^5-8=2y^2$$

if you introduce $$y=2^x$$.

• but you need a software (calculator) to solve that, correct? – bru1987 Oct 5 '18 at 13:03
• @bru1987 Yes, the solution to this equation can only be calculated numerically, but there's really no going around this fact. Any solution of your original equation would also give rise to a solution of this equation, after all. – 5xum Oct 5 '18 at 13:11
• thank you for the answer. – bru1987 Oct 5 '18 at 13:36