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Question Number 97725 by john santu last updated on 09/Jun/20

$$\log {}_5\left(4\mathrm{x}\right)=\log {}_{10}\left(\mathrm{x}\right)$$$$\mathrm{find}\mathrm{x}?$$

Commented by prakash jain last updated on 09/Jun/20

$$4x=5^y$$$$x={10}^y$$$$4\times {10}^y=5^y$$$$2^y=\frac{1}{4}\Rightarrow y=-2$$$$x=\frac{1}{100}$$

Commented by bobhans last updated on 09/Jun/20

$$\frac{\ln \left(4\mathrm{x}\right)}{\ln \left(5\right)}=\frac{\ln \left(\mathrm{x}\right)}{\ln \left(10\right)}$$$$\frac{2\ln \left(2\right)+\ln \left(\mathrm{x}\right)}{\ln \left(5\right)}=\frac{\ln \left(\mathrm{x}\right)}{\ln \left(5\right)+\ln \left(2\right)}$$$$\ln \left(5\right)\ln \left(\mathrm{x}\right)=(2\ln \left(2\right)+\ln \left(\mathrm{x}\right))(\ln \left(5\right)+\ln \left(2\right))$$$$\ln \left(5\right)\ln \left(\mathrm{x}\right)=2\ln \left(5\right)\ln \left(2\right)+{2\ln }^2\left(2\right)+\ln \left(5\right)\ln \left(\mathrm{x}\right)+\ln \left(\mathrm{x}\right)\ln \left(2\right)$$$$0=2\ln \left(5\right)\ln \left(2\right)+{2\ln }^2\left(2\right)+\ln \left(\mathrm{x}\right)\ln \left(2\right)$$$$0=2\ln \left(5\right)+2\ln \left(2\right)+\ln \left(\mathrm{x}\right)$$$$\ln \left(\mathrm{x}\right)=-2\ln \left(10\right)=\ln \left({10}^{-2}\right)$$$$\iff \mathrm{x}=\frac{1}{100}$$

Commented by john santu last updated on 09/Jun/20

$$\mathrm{thank}\mathrm{both}$$

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