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Question Number 196946 by Khalmohmmad last updated on 04/Sep/23

Answered by trisetyo last updated on 04/Sep/23

$$3^x+2^x=5$$$$3^x=-(2^x-5)\vee 2^x=-(3^x-5)$$$${\log }_3(-2^x+5)=x\mathrm{and}{\log }_2(-3^x+5)=x$$$$\frac{\log (-2^x+5)}{\log \left(3\right)}=\frac{\log (-3^x+5)}{\log \left(2\right)}$$$$\frac{\log (-2^x+5)}{\log (-3^x+5)}=\frac{\log \left(3\right)}{\log \left(2\right)}$$$$\Rightarrow -2^x+5=3\vee -3^x+5=2$$$$\Rightarrow 2^x=2\vee 3^x=3$$$$\Rightarrow x={\log }_2\left(2\right)\vee x={\log }_3\left(3\right)$$$$\Rightarrow x=1\vee x=1$$$$\therefore x=1$$

Commented by Frix last updated on 04/Sep/23

$$\mathrm{Nonsense}.$$$$\mathrm{At}\mathrm{least}\mathrm{test}\mathrm{your}\mathrm{solution}$$$$3^x-2^x=3^1-2^1=1\ne 5$$

Commented by trisetyo last updated on 04/Sep/23

$$$$$$\mathrm{oh}{\mathrm{i}}^{\prime }\mathrm{m}\mathrm{sorry}...\mathrm{its}{3}^x-2^x=5$$$$\mathrm{my}\mathrm{answer}{3}^x+2^x=5$$$$\mathrm{it}\mathrm{was}\mathrm{my}\mathrm{mistake},sorry..$$

Answered by Frix last updated on 04/Sep/23

$$3^2-2^2=9-4=5$$$$x=2$$

Commented by trisetyo last updated on 04/Sep/23

$$$$$$verrysimplesolutionXD$$

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