2
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Any neat way to solve this equation?

NSolve[(8^x - 2^x)/(6^x - 3^x) == 2, x]

NSolve[(8^x - 2^x)/(6^x - 3^x) == 2, x, 
 Method -> {"UseSlicingHyperplanes" -> False}]

Output says: This system cannot be solved with the methods available to NSolve.

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2
  • 1
    $\begingroup$ FindInstance[(8^x - 2^x)/(6^x - 3^x) == 2 && x > 0, x, Reals, 1] $\endgroup$
    – I.M.
    Dec 1, 2022 at 5:24
  • 1
    $\begingroup$ NSolve[(8^x - 2^x)/(6^x - 3^x) == 2, Reals] $\endgroup$
    – cvgmt
    Dec 1, 2022 at 8:16

2 Answers 2

3
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 Plot[{(8^x - 2^x)/(6^x - 3^x), 2}, {x, -3, 3}]

Mathematica graphics

So you can try to help NSolve a little

 NSolve[(8^x - 2^x)/(6^x - 3^x) == 2 && 0 <= x , x]

Mathematica graphics

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-1
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NSolve[(8^x - 2^x)/(6^x - 3^x) == 2 && x > -10 && x <= 10, x, Reals]

{{x -> 1.}}

FindInstance[(8^x-2^x)/(6^x-3^x)==2&&x>-10&&x<=10,x,Reals]

{{x -> 1}}

Unfortunately,

FindInstance[(8^x - 2^x)/(6^x - 3^x) == 2 && x > -10 &&   x <= 10, x, Reals, 2]

{{x -> 0}, {x -> 1}}

which is not correct.

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1
  • 1
    $\begingroup$ Since Limit[(8^x - 2^x)/(6^x - 3^x), x -> 0]==Log[4]/Log[2], one can redefine (8^x - 2^x)/(6^x - 3^x) at x==0 to make the result of FindInstance correct. – $\endgroup$
    – user64494
    Dec 1, 2022 at 5:56

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