Question Number 122286 by ZiYangLee last updated on 15/Nov/20
$$\left(\mathrm{49}^{\mathrm{log}_{\mathrm{7}} \mathrm{6}} \right)^{\mathrm{log}_{\mathrm{3}} \mathrm{5}^{\mathrm{log}_{\mathrm{2}} \mathrm{3}^{\mathrm{log}_{\mathrm{5}} \mathrm{2}} } } =? \\ $$
Answered by TANMAY PANACEA last updated on 15/Nov/20
$$ \\ $$$$\left(\mathrm{49}^{{log}_{\mathrm{7}} ^{\mathrm{6}} } \right)^{{log}_{\mathrm{3}} \mathrm{5}^{{log}_{\mathrm{2}\:} \mathrm{3}^{{log}_{\mathrm{5}\:} \mathrm{2}} } } \\ $$$$\left(\mathrm{49}^{{a}} \right)^{{b}^{\frac{{ln}\mathrm{2}}{{ln}\mathrm{5}}×\frac{{ln}\mathrm{3}}{{ln}\mathrm{2}}} } \\ $$$$\left(\mathrm{49}^{{a}} \right)^{{b}^{\frac{{ln}\mathrm{3}}{{ln}\mathrm{5}}} } \\ $$$$\left(\mathrm{49}^{{a}} \right)^{\frac{{ln}\mathrm{5}}{{ln}\mathrm{3}}×\frac{{ln}\mathrm{3}}{{ln}\mathrm{5}}} \\ $$$$\mathrm{49}^{{a}} ={p} \\ $$$${ln}_{\mathrm{49}} {p}={a} \\ $$$$\frac{{lnp}}{\mathrm{2}{ln}\mathrm{7}}={a}=\frac{{ln}\mathrm{6}}{{ln}\mathrm{7}} \\ $$$${lnp}=\mathrm{2}{ln}\mathrm{6}={ln}\mathrm{6}^{\mathrm{2}} \\ $$$${p}=\mathrm{36} \\ $$