If-ax-loga-bx-logb-then-prove-that-x-1-ab-

Question Number 196690 by MATHEMATICSAM last updated on 29/Aug/23

$$\mathrm{If}\:\left({ax}\right)^{\mathrm{log}{a}} \:=\:\left({bx}\right)^{\mathrm{log}{b}} \:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$${x}\:=\:\frac{\mathrm{1}}{{ab}}\:. \\ $$

Answered by BaliramKumar last updated on 29/Aug/23

loga∙log(ax) = logb∙log(bx) loga∙(loga + logx) = logb∙(logb + logx) log^2 a + loga∙ logx = log^2 b + logb∙logx loga∙ logx − logb∙logx = log^2 b − log^2 a logx(loga − logb) = (logb − loga)(logb + loga) logx∙log((a/b)) = −log((a/b))∙logab logx = −log(ab) logx = log((1/(ab))) x = (1/(ab))

$$\mathrm{loga}\centerdot\mathrm{log}\left(\mathrm{ax}\right)\:=\:\mathrm{logb}\centerdot\mathrm{log}\left(\mathrm{bx}\right) \\ $$$$\mathrm{loga}\centerdot\left(\mathrm{loga}\:+\:\mathrm{logx}\right)\:=\:\mathrm{logb}\centerdot\left(\mathrm{logb}\:+\:\mathrm{logx}\right) \\ $$$$\mathrm{log}^{\mathrm{2}} \mathrm{a}\:+\:\mathrm{loga}\centerdot\:\mathrm{logx}\:=\:\mathrm{log}^{\mathrm{2}} \mathrm{b}\:+\:\mathrm{logb}\centerdot\mathrm{logx} \\ $$$$\mathrm{loga}\centerdot\:\mathrm{logx}\:−\:\:\mathrm{logb}\centerdot\mathrm{logx}\:=\:\mathrm{log}^{\mathrm{2}} \mathrm{b}\:−\:\mathrm{log}^{\mathrm{2}} \mathrm{a} \\ $$$$\mathrm{logx}\left(\mathrm{loga}\:−\:\mathrm{logb}\right)\:=\:\left(\mathrm{logb}\:−\:\mathrm{loga}\right)\left(\mathrm{logb}\:+\:\mathrm{loga}\right) \\ $$$$\mathrm{logx}\centerdot\cancel{\mathrm{log}\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}\:=\:−\cancel{\mathrm{log}\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}\centerdot\mathrm{logab} \\ $$$$\mathrm{logx}\:=\:−\mathrm{log}\left(\mathrm{ab}\right) \\ $$$$\mathrm{logx}\:=\:\mathrm{log}\left(\frac{\mathrm{1}}{\mathrm{ab}}\right) \\ $$$$\mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{ab}} \\ $$

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