Question Number 221869 by fantastic last updated on 12/Jun/25
$${if}\:{a}^{\mathrm{3}−{x}} .{b}^{\mathrm{5}{x}} ={a}^{\mathrm{5}+{x}} .{b}^{\mathrm{3}{x}} \:{then}\:{show}\:{that} \\ $$$${x}\mathrm{log}\:\left(\frac{{b}}{{a}}\right)=\mathrm{log}\:{a} \\ $$
Answered by fantastic last updated on 12/Jun/25
$${we}\:{can}\:{write}\:\frac{{a}^{\mathrm{3}} }{{a}^{{x}} }.{b}^{\mathrm{5}{x}} ={a}^{\mathrm{5}} .{a}^{{x}} .{b}^{\mathrm{3}{x}} \\ $$$${or}\:{b}^{\mathrm{2}{x}} ={a}^{\mathrm{2}{x}} .{a}^{\mathrm{2}} \\ $$$${or}\:{b}^{{x}} ={a}^{{x}} .{a} \\ $$$${or}\:\left(\frac{{b}}{{a}}\right)^{{x}} ={a} \\ $$$${So}\:\mathrm{log}\:\left(\frac{{b}}{{a}}\right)^{{x}} =\mathrm{log}\:{a} \\ $$$$\therefore\:{x}\mathrm{log}\:\left(\frac{{b}}{{a}}\right)=\mathrm{log}\:{a} \\ $$
Commented by Rasheed.Sindhi last updated on 12/Jun/25
$$\mathbb{N}\boldsymbol{\mathrm{i}}\subset\in! \\ $$
Commented by fantastic last updated on 12/Jun/25
$$\tau{h}\alpha\eta\kappa\Im \\ $$