Question Number 210034 by peter frank last updated on 29/Jul/24
Commented by peter frank last updated on 29/Jul/24
$$\mathrm{solve}\:\mathrm{for}\:\mathrm{x} \\ $$
Commented by Frix last updated on 29/Jul/24
$$\left(\mathrm{log}\:\mathrm{5}\right)^{\mathrm{log}\:{x}} \:\mathrm{or}\:\mathrm{log}\:\left(\mathrm{5}^{\mathrm{log}\:{x}} \right)? \\ $$$$\mathrm{log}=\mathrm{ln}\:\mathrm{or}\:\mathrm{log}=\mathrm{log}_{\mathrm{10}} ? \\ $$
Commented by peter frank last updated on 29/Jul/24
$$\mathrm{log}\left(\mathrm{5}^{\mathrm{logx}} \right) \\ $$
Commented by Frix last updated on 29/Jul/24
$$\mathrm{log}\:\left(\mathrm{5}^{\mathrm{log}\:{x}} \right)\:=\mathrm{log}\:{x}\:×\mathrm{log}\:\mathrm{5} \\ $$$$\mathrm{For}\:\mathrm{the}\:\mathrm{base}\:{b}: \\ $$$$\mathrm{log}_{{b}} \:{x}\:\frac{\mathrm{4}}{\mathrm{log}_{{b}} \:\mathrm{5}} \\ $$$${x}={b}^{\frac{\mathrm{4}}{\mathrm{log}_{{b}} \:\mathrm{5}}} \\ $$$${b}=\mathrm{10}\:\Rightarrow\:{x}=\mathrm{10}^{\frac{\mathrm{4}}{\mathrm{log}_{\mathrm{10}} \:\mathrm{5}}} \approx\mathrm{528087}.\mathrm{919} \\ $$$${b}=\mathrm{e}\:\Rightarrow\:{x}=\mathrm{e}^{\frac{\mathrm{4}}{\mathrm{ln}\:\mathrm{5}}} \approx\mathrm{12}.\mathrm{0051983} \\ $$
Answered by Spillover last updated on 29/Jul/24
