Question Number 52091 by Olalekan99 last updated on 03/Jan/19
$$\mathrm{3}^{{x}} +\mathrm{4}^{{x}} =\mathrm{5}^{{x}} \\ $$$${find}\:{x} \\ $$$${BY}\:{ISHOLA} \\ $$
Commented by scientist last updated on 03/Jan/19
$${using}\:{binomial}\:{expansion}\: \\ $$$$\left(\mathrm{1}+\mathrm{2}\right)^{{x}} +\left(\mathrm{1}+\mathrm{3}\right)^{{x}} =\left(\mathrm{1}+\mathrm{4}\right)^{{x}} \\ $$
Answered by scientist last updated on 03/Jan/19
$${x}=\mathrm{2} \\ $$
Answered by mr W last updated on 03/Jan/19
$${f}\left({x}\right)=\mathrm{3}^{{x}} +\mathrm{4}^{{x}} \:{is}\:{increasing}\:{function} \\ $$$${g}\left({x}\right)=\mathrm{5}^{{x}} \:{is}\:{increasing}\:{function} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{3}+\mathrm{4}=\mathrm{7} \\ $$$${g}\left(\mathrm{1}\right)=\mathrm{5} \\ $$$${f}\left(\mathrm{1}\right)>{g}\left(\mathrm{1}\right) \\ $$$${f}\left(\mathrm{3}\right)=\mathrm{3}^{\mathrm{3}} +\mathrm{4}^{\mathrm{3}} =\mathrm{91} \\ $$$${g}\left(\mathrm{3}\right)=\mathrm{5}^{\mathrm{3}} =\mathrm{125} \\ $$$${f}\left(\mathrm{3}\right)<{g}\left(\mathrm{3}\right) \\ $$$$\Rightarrow{there}\:{is}\:{one}\:{and}\:{only}\:{one}\:{solution}\:{between} \\ $$$$\mathrm{1}\:{and}\:\mathrm{3}\:{for}\:{f}\left({x}\right)={g}\left({x}\right). \\ $$$${since}\:{f}\left(\mathrm{2}\right)=\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} =\mathrm{25}=\mathrm{5}^{\mathrm{2}} ={g}\left({x}\right), \\ $$$${the}\:{only}\:{solution}\:{is}\:{x}=\mathrm{2}. \\ $$