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Question Number 170958 by cortano1 last updated on 04/Jun/22

$$\log {}_5(x+3)=\log {}_6(x+14)$$

Answered by floor(10²Eta[1]) last updated on 05/Jun/22

$${\log }_5(\mathrm{x}+3)=\mathrm{y}={\log }_6(\mathrm{x}+14)$$$$\mathrm{x}+3=5^{\mathrm{y}}$$$$\mathrm{x}+14=6^{\mathrm{y}}$$$$5^{\mathrm{y}}+11=6^{\mathrm{y}}$$$$\mathrm{y}=2\Rightarrow 5^2+11=6^2$$$${\mathrm{let}}^{\prime }\mathrm{s}\mathrm{show}\mathrm{that}\mathrm{for}\mathrm{y}>2\Rightarrow 6^{\mathrm{y}}>5^{\mathrm{y}}+11$$$$\mathrm{let}\mathrm{f}\left(\mathrm{y}\right)=6^{\mathrm{y}}-5^{\mathrm{y}}-11$$$$\mathrm{we}\mathrm{want}:\mathrm{f}\left(\mathrm{y}\right)>0,\mathrm{\forall }\mathrm{y}>2$$$$\mathrm{f}\left(2\right)=0$$$${\mathrm{f}}^{\prime }\left(\mathrm{y}\right)=6^{\mathrm{y}}\ln 6-5^{\mathrm{y}}\ln 5$$$$\mathrm{but}{\mathrm{f}}^{\prime }\left(\mathrm{y}\right)>0,\mathrm{\forall }\mathrm{y}>2$$$$\Rightarrow \mathrm{f}\mathrm{is}\mathrm{increasing}\mathrm{on}\mathrm{the}\mathrm{interval}[2,+\mathrm{\infty })$$$$\mathrm{since}\mathrm{f}\left(2\right)=0\Rightarrow \mathrm{f}\left(\mathrm{y}\right)>0,\mathrm{\forall }\mathrm{y}>2$$$$\Rightarrow 6^{\mathrm{y}}>5^{\mathrm{y}}+11,\mathrm{\forall }\mathrm{y}>2$$$$\Rightarrow \mathrm{y}=2\mathrm{is}\mathrm{the}\mathrm{only}\mathrm{solution}\mathrm{to}{6}^{\mathrm{y}}=5^{\mathrm{y}}+11$$$$\Rightarrow \mathrm{x}+3=5^2\Rightarrow \mathrm{x}=22$$

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