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Question Number 142755 by ERA last updated on 05/Jun/21

$${\log }_3{\mathrm{x}}^3+{\log }_2{\mathrm{x}}^2=\frac{2\lg 6}{\lg 2}+1\mathrm{find}\mathrm{x}$$

Commented by iloveisrael last updated on 05/Jun/21

$$whatdifferent\log and\lg ?$$

Commented by qaz last updated on 05/Jun/21

$$\mathrm{lgx}={\log }_{10}\mathrm{x}$$

Answered by iloveisrael last updated on 05/Jun/21

$$\Rightarrow \log {}_3\left(x^3\right)+\log {}_2\left(x^2\right)=2\log {}_2\left(3\times 2\right)+\log {}_3\left(3\right)$$$$\Rightarrow 3\log {}_3\left(x\right)+2\log {}_2\left(x\right)=2\log {}_2\left(2\right)+3\log {}_3\left(3\right)$$$$\Rightarrow {3\log }_3\left(\frac{x}{3}\right)={2\log }_2\left(\frac{2}{x}\right)$$$$\Rightarrow {3\log }_3\left(\frac{x}{3}\right)=-{2\log }_2\left(\frac{x}{2}\right)$$$$\Rightarrow \frac{3\ln x-3\ln 3}{\ln 3}=\frac{-2\ln x+2\ln 2}{\ln 2}$$$$\Rightarrow 3\ln 2.\ln x-3\ln 2.\ln 3=-2\ln 3.\ln x+2\ln 2.\ln 3$$$$\Rightarrow 2\ln x.\cancel{\ln 8}=2\ln 2.\cancel{\ln 8}$$$$\Rightarrow x=2$$

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