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Question Number 93847 by i jagooll last updated on 15/May/20

$$\log {}_5(\mathrm{x}-2)+\log {}_8(\mathrm{x}-4)=\log {}_6(\mathrm{x}-1)$$

Commented by john santu last updated on 15/May/20

$$$$$$\frac{\ln (\mathrm{x}-2)}{\ln 5}+\frac{\ln (\mathrm{x}-4)}{\ln 8}=\frac{\ln (\mathrm{x}-1)}{\ln 6}$$$$\mathrm{x}>4\mathrm{for}\mathrm{x}\in \mathbb{R}$$$$\ln {(\mathrm{x}-2)}^{0.62}+\ln {(\mathrm{x}-4)}^{0.48}=\ln {(\mathrm{x}-1)}^{0.59}$$$$\ln {(\mathrm{x}-2)}^{0.62}.{(\mathrm{x}-4)}^{0.48}=\ln {(\mathrm{x}-1)}^{0.59}$$$${(\mathrm{x}-2)}^{0.62}.{(\mathrm{x}-4)}^{0.48}={(\mathrm{x}-1)}^{0.59}$$

Commented by i jagooll last updated on 15/May/20

$$\mathbb{W}\mathrm{hat}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{x}\mathrm{sir}?$$

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