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Question Number 76034 by hmamarques1994@gmail.com last updated on 22/Dec/19

53^(log_x (7)) = (√x) x = ?

$$\: \\ $$$$\:\mathrm{53}^{\boldsymbol{\mathrm{log}}_{\boldsymbol{\mathrm{x}}} \left(\mathrm{7}\right)} \:=\:\sqrt{\boldsymbol{\mathrm{x}}} \\ $$$$\: \\ $$$$\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$$$\: \\ $$

Answered by MJS last updated on 22/Dec/19

53^((ln 7)/(ln x)) =(√x) ((ln 7)/(ln x))ln 53 =ln (√x) ((ln 7 ln 53)/(ln x))=((ln x)/2) (ln x)^2 =2ln 7 ln 53 ln x =±(√(2 ln 7 ln 53)) x=e^(±(√(2 ln 7 ln 53))) x≈.0196268∨x≈50.9508

$$\mathrm{53}^{\frac{\mathrm{ln}\:\mathrm{7}}{\mathrm{ln}\:{x}}} =\sqrt{{x}} \\ $$$$\frac{\mathrm{ln}\:\mathrm{7}}{\mathrm{ln}\:{x}}\mathrm{ln}\:\mathrm{53}\:=\mathrm{ln}\:\sqrt{{x}} \\ $$$$\frac{\mathrm{ln}\:\mathrm{7}\:\mathrm{ln}\:\mathrm{53}}{\mathrm{ln}\:{x}}=\frac{\mathrm{ln}\:{x}}{\mathrm{2}} \\ $$$$\left(\mathrm{ln}\:{x}\right)^{\mathrm{2}} =\mathrm{2ln}\:\mathrm{7}\:\mathrm{ln}\:\mathrm{53} \\ $$$$\mathrm{ln}\:{x}\:=\pm\sqrt{\mathrm{2}\:\mathrm{ln}\:\mathrm{7}\:\mathrm{ln}\:\mathrm{53}} \\ $$$${x}=\mathrm{e}^{\pm\sqrt{\mathrm{2}\:\mathrm{ln}\:\mathrm{7}\:\mathrm{ln}\:\mathrm{53}}} \\ $$$${x}\approx.\mathrm{0196268}\vee{x}\approx\mathrm{50}.\mathrm{9508} \\ $$

Commented by David Danile last updated on 04/Jan/20

can′t we solve this using the fact tbat b^(log_x a^c ) = c ?

$${can}'{t}\:{we}\:{solve}\:{this}\:{using}\:{the}\:{fact}\:{tbat}\: \\ $$$$ \\ $$$${b}^{{log}_{{x}} {a}^{{c}} } =\:{c}\:\:? \\ $$

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