log-8-log-2-log-3-4-x-17-1-3-x-

Question Number 223514 by fantastic last updated on 27/Jul/25

$$\mathrm{log}\:_{\mathrm{8}} \left[\mathrm{log}\:_{\mathrm{2}} \left\{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{4}^{{x}} +\mathrm{17}\right)\right\}\right]=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${x}=?? \\ $$

Answered by som(math1967) last updated on 27/Jul/25

$${log}_{\mathrm{2}} \left\{{log}_{\mathrm{3}} \left(\mathrm{4}^{{x}} +\mathrm{17}\right)\right\}=\mathrm{8}^{\frac{\mathrm{1}}{\mathrm{3}}} =\mathrm{2} \\ $$$${log}_{\mathrm{3}} \left(\mathrm{4}^{{x}} +\mathrm{17}\right)=\mathrm{2}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{4}^{{x}} +\mathrm{17}=\mathrm{3}^{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{4}^{{x}} =\mathrm{81}−\mathrm{17} \\ $$$$\Rightarrow\mathrm{4}^{{x}} =\mathrm{4}^{\mathrm{3}} \:\:\:\:\:\:\therefore{x}=\mathrm{3} \\ $$

Commented by fantastic last updated on 27/Jul/25

$${thanks} \\ $$

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