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Question Number 198222 by cortano12 last updated on 14/Oct/23

$$\log {}_4(5^{\mathrm{x}}-3^{\mathrm{x}})=\log {}_5(4^{\mathrm{x}}+3^{\mathrm{x}})$$

Commented by Rasheed.Sindhi last updated on 14/Oct/23

$$4^2+3^2=5^2$$$${\log }_5(4^2+3^2)={\log }_5{5}^2$$$${\log }_5(4^2+3^2)=2......\left(i\right)$$$$$$$$5^2-3^2=4^2$$$${\log }_4(5^2-3^2)={\log }_4{4}^2$$$${\log }_4(5^2-3^2)=2......\left(ii\right)$$$$\left(i\right)\mathrm{\&}\left(ii\right):$$$${\log }_5(4^2+3^2)={\log }_4(5^2-3^2)$$$$\mathrm{Comparing}\mathrm{with}$$$${\log }_5(4^{\mathrm{x}}+3^{\mathrm{x}})={\log }_4(5^{\mathrm{x}}-3^{\mathrm{x}})$$$$\mathrm{yields}$$$$\mathrm{x}=2$$

Answered by Rasheed.Sindhi last updated on 14/Oct/23

$$\log {}_4(5^{\mathrm{x}}-3^{\mathrm{x}})=\log {}_5(4^{\mathrm{x}}+3^{\mathrm{x}})=\mathrm{k}$$$$4^{\mathrm{k}}=5^{\mathrm{x}}-3^{\mathrm{x}},5^{\mathrm{k}}=4^{\mathrm{x}}+3^{\mathrm{x}}$$$$4^{\mathrm{k}}+5^{\mathrm{k}}=5^{\mathrm{x}}+4^{\mathrm{x}}\Rightarrow \mathrm{k}=\mathrm{x}$$$$\therefore \log {}_4(5^{\mathrm{x}}-3^{\mathrm{x}})=\log {}_5(4^{\mathrm{x}}+3^{\mathrm{x}})=\mathrm{x}$$$$4^{\mathrm{x}}=5^{\mathrm{x}}-3^{\mathrm{x}},5^{\mathrm{x}}=4^{\mathrm{x}}+3^{\mathrm{x}}$$$$\mathrm{Both}\mathrm{are}\mathrm{equivalent}\mathrm{to}:$$$$4^{\mathrm{x}}+3^{\mathrm{x}}=5^{\mathrm{x}}$$$$\mathrm{Which}\mathrm{is}\mathrm{only}\mathrm{true}\mathrm{for}$$$$\mathrm{x}=2$$

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