Question Number 53827 by Mikael_Marshall last updated on 26/Jan/19
$$\mathrm{5}×\mathrm{6}^{{x}} −\mathrm{3}×\mathrm{4}^{{x}} =\:\mathrm{2}×\mathrm{9}^{{x}} \\ $$$$ \\ $$$${I}\:{need}\:{an}\:{explanation}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 26/Jan/19
$$\mathrm{5}×\mathrm{3}^{{x}} ×\mathrm{2}^{{x}} −\mathrm{3}×\mathrm{2}^{{x}} ×\mathrm{2}^{{x}} =\mathrm{2}×\mathrm{3}^{{x}} ×\mathrm{3}^{{x}} \\ $$$${a}=\mathrm{2}^{{x}} \:\:\:{b}=\mathrm{3}^{{x}} \\ $$$$\mathrm{5}{ab}−\mathrm{3}{a}^{\mathrm{2}} =\mathrm{2}{b}^{\mathrm{2}} \\ $$$$\mathrm{3}{a}^{\mathrm{2}} −\mathrm{5}{ab}+\mathrm{2}{b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{3}{a}^{\mathrm{2}} −\mathrm{3}{ab}−\mathrm{2}{ab}+\mathrm{2}{b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{3}{a}\left({a}−{b}\right)−\mathrm{2}{b}\left({a}−{b}\right)=\mathrm{0} \\ $$$$\left({a}−{b}\right)\left(\mathrm{3}{a}−\mathrm{2}{b}\right)=\mathrm{0} \\ $$$${either}\:{a}−{b}=\mathrm{0} \\ $$$${a}={b} \\ $$$$\mathrm{2}^{{x}} =\mathrm{3}^{{x}} \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} =\mathrm{1}=\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{0}} \:\:{so}\:{x}=\mathrm{0} \\ $$$${or}\:{when} \\ $$$$\mathrm{3}{a}−\mathrm{2}{b}=\mathrm{0} \\ $$$$\mathrm{3}×\mathrm{2}^{{x}} =\mathrm{2}×\mathrm{3}^{{x}} \\ $$$$\frac{\mathrm{2}^{{x}} }{\mathrm{3}^{{x}} }=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} =\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{1}} \:\:\:{so}\:{x}=\mathrm{1} \\ $$$${hence}\:{x}\:{is}\:{either}\:\mathrm{0}\:{or}\:\mathrm{1} \\ $$
Commented by Mikael_Marshall last updated on 26/Jan/19
$${thank}\:\:{you}\:{bro} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 26/Jan/19
$${most}\:{welcome}... \\ $$