Question Number 59730 by muneshkumar last updated on 14/May/19
$$\mathrm{5}^{\mathrm{2}{x}−\mathrm{1}\:} =\mathrm{25}^{{x}−\mathrm{1}} +\mathrm{100}\:{find}\:{value}\:{of}\:\mathrm{3}^{\mathrm{3}−{x}} \\ $$
Answered by tanmay last updated on 14/May/19
$$\frac{\mathrm{5}^{\mathrm{2}{x}} }{\mathrm{5}}=\left(\mathrm{5}^{\mathrm{2}} \right)^{{x}−\mathrm{1}} +\mathrm{100} \\ $$$$\frac{\left(\mathrm{5}^{{x}} \right)^{\mathrm{2}} }{\mathrm{5}}=\frac{\mathrm{5}^{\mathrm{2}{x}} }{\mathrm{5}^{\mathrm{2}} }+\mathrm{100} \\ $$$$\mathrm{5}^{{x}} ={a} \\ $$$$\frac{{a}^{\mathrm{2}} }{\mathrm{5}}=\frac{{a}^{\mathrm{2}} }{\mathrm{25}}+\mathrm{100} \\ $$$$\frac{{a}^{\mathrm{2}} }{\mathrm{5}}−\frac{{a}^{\mathrm{2}} }{\mathrm{25}}=\mathrm{100} \\ $$$$\frac{\mathrm{5}{a}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{25}}=\mathrm{100} \\ $$$$\mathrm{4}{a}^{\mathrm{2}} =\mathrm{100}×\mathrm{25} \\ $$$${a}^{\mathrm{2}} =\left(\frac{\mathrm{10}×\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{5}^{\mathrm{4}} \\ $$$$\mathrm{5}^{\mathrm{2}{x}} =\mathrm{5}^{\mathrm{4}} \\ $$$$\mathrm{2}{x}=\mathrm{4} \\ $$$${x}=\mathrm{2} \\ $$$${so} \\ $$$$\mathrm{3}^{\mathrm{3}−\mathrm{2}} \\ $$$$\mathrm{3}^{\mathrm{3}−\mathrm{2}} \\ $$$$\mathrm{3}^{\mathrm{1}} \\ $$$$\mathrm{3} \\ $$$$ \\ $$$$ \\ $$