1010-x-2020-x-4040-x-x-

Question Number 96289 by rb222 last updated on 31/May/20

1010^{x} + 2020^{x} = 4040^{x}

x = ?

Commented by bobhans last updated on 31/May/20

{(\frac{1}{4})}^{x} + {(\frac{1}{2})}^{x} = 1 [t = {(\frac{1}{2})}^{x}]

t^{2} + t - 1 = 0 \Rightarrow t = \frac{- 1 + \sqrt{5}}{2}

- x \ln (2) = \ln (\sqrt{5} - 1) - \ln (2)

x = \frac{\ln (2) - \ln (\sqrt{5} - 1)}{\ln (2)}

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