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Question Number 111721 by Aina Samuel Temidayo last updated on 04/Sep/20

$$\mathrm{If}\mathrm{b}>1,\mathrm{x}>0\mathrm{and}{\left(2\mathrm{x}\right)}^{{\log }_{\mathrm{b}}2}-{\left(3\mathrm{x}\right)}^{{\log }_{\mathrm{b}}3}=0,$$ $$\mathrm{then}\mathrm{x}\mathrm{is}$$

Answered by Her_Majesty last updated on 05/Sep/20

$${\left(2x\right)}^{{log}_{b}2}={\left(3x\right)}^{{log}_{b}3}$$ $${log}_{b}2\times ln\left(2x\right)={log}_{b}3\times ln\left(3x\right)$$ $$\frac{ln2}{lnb}ln\left(2x\right)=\frac{ln3}{lnb}ln\left(3x\right)$$ $$ln2(ln2+lnx)=ln3(ln3+lnx)$$ $${\left(ln2\right)}^2+ln2lnx={\left(ln3\right)}^2+ln3lnx$$ $$(ln3-ln2)lnx={\left(ln2\right)}^2-{\left(ln3\right)}^2$$ $$lnx=\frac{(ln3-ln3)(ln2+ln3)}{ln3-ln2}$$ $$lnx=-(ln2+ln3)$$ $$lnx=-ln6$$ $$lnx=ln\frac{1}{6}$$ $$x=\frac{1}{6}$$

Commented byAina Samuel Temidayo last updated on 05/Sep/20

$$\mathrm{Thanks}.$$

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