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Question Number 209341 by Tawa11 last updated on 07/Jul/24

$$\mathrm{solve}{\mathrm{x}}^{\log 27}+9^{\log \mathrm{x}}=36$$

Answered by Frix last updated on 07/Jul/24

$$x^{{\log }_{b}27}=x^{\frac{3\ln 3}{\ln b}}$$$$9^{{\log }_{b}x}=x^{\frac{2\ln 3}{\mathrm{kn}b}}$$$$x=t^{\frac{\ln b}{\ln 3}}$$$$\Rightarrow x^{{\log }_{b}27}+9^{{\log }_{b}x}=t^3+t^2=36=27+9\Rightarrow t=3$$$$\Rightarrow$$$$x=b$$$$\mathrm{If}‘‘{\log }^{″}={\log }_{10}\Rightarrow x=10$$$$\mathrm{If}‘‘{\log }^{″}=\ln \Rightarrow x=\mathrm{e}$$

Commented by Tawa11 last updated on 07/Jul/24

$$\mathrm{Thanks}\mathrm{sir}$$

Answered by Kademi last updated on 09/Jul/24

$$$$$$x^{\log 27}+9^{\log x}=36$$$${27}^{\log x}+9^{\log x}={27}^1+9^1$$$$\Rightarrow \log x=1\iff x=10$$

Commented by Tawa11 last updated on 09/Jul/24

$$\mathrm{Thanks}\mathrm{sir}$$

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