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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.

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    $\begingroup$ This leads to a difficult polynomial equation. Are you sure about the coefficients ? $\endgroup$
    – user65203
    Oct 5, 2018 at 12:46
  • $\begingroup$ @YvesDaoust yes, exactly my conclusion. Yes I am sure. $\endgroup$
    – bru1987
    Oct 5, 2018 at 13:01
  • $\begingroup$ Where is that coming from ? $\endgroup$
    – user65203
    Oct 5, 2018 at 13:06
  • $\begingroup$ From a colleague's material. He had that on a 10th grade, no calc, test. $\endgroup$
    – bru1987
    Oct 5, 2018 at 13:07
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    $\begingroup$ I'm thinking so too, the only real solution isn't a rational root as found by wolfram. $\endgroup$ Oct 5, 2018 at 13:27

1 Answer 1

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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 $y=2^x$.

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    $\begingroup$ but you need a software (calculator) to solve that, correct? $\endgroup$
    – bru1987
    Oct 5, 2018 at 13:03
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    $\begingroup$ @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. $\endgroup$
    – 5xum
    Oct 5, 2018 at 13:11
  • $\begingroup$ thank you for the answer. $\endgroup$
    – bru1987
    Oct 5, 2018 at 13:36

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