Question Number 96289 by rb222 last updated on 31/May/20
$$\mathrm{1010}^{{x}} +\mathrm{2020}^{{x}} =\mathrm{4040}^{{x}} \\ $$$${x}=? \\ $$
Commented by bobhans last updated on 31/May/20
![((1/4))^x +((1/2))^x =1 [ t = ((1/2))^x ] t^2 +t−1= 0 ⇒ t = ((−1+ (√5))/2) −x ln(2) = ln((√5)−1)−ln(2) x = ((ln(2)−ln((√5)−1))/(ln(2)))](Q96291.png)
$$\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{{x}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{x}} =\mathrm{1}\:\left[\:\mathrm{t}\:=\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{x}} \right] \\ $$$${t}^{\mathrm{2}} +{t}−\mathrm{1}=\:\mathrm{0}\:\Rightarrow\:\mathrm{t}\:=\:\frac{−\mathrm{1}+\:\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$−{x}\:\mathrm{ln}\left(\mathrm{2}\right)\:=\:\mathrm{ln}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)−\mathrm{ln}\left(\mathrm{2}\right)\: \\ $$$${x}\:=\:\frac{\mathrm{ln}\left(\mathrm{2}\right)−\mathrm{ln}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)}{\mathrm{ln}\left(\mathrm{2}\right)} \\ $$