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49-log-7-6-log-3-5-log-2-3-log-5-2-




Question Number 122286 by ZiYangLee last updated on 15/Nov/20
(49^(log_7 6) )^(log_3 5^(log_2 3^(log_5 2) ) ) =?
$$\left(\mathrm{49}^{\mathrm{log}_{\mathrm{7}} \mathrm{6}} \right)^{\mathrm{log}_{\mathrm{3}} \mathrm{5}^{\mathrm{log}_{\mathrm{2}} \mathrm{3}^{\mathrm{log}_{\mathrm{5}} \mathrm{2}} } } =? \\ $$
Answered by TANMAY PANACEA last updated on 15/Nov/20
  (49^(log_7 ^6 ) )^(log_3 5^(log_(2 ) 3^(log_(5 ) 2) ) )   (49^a )^b^(((ln2)/(ln5))×((ln3)/(ln2)))    (49^a )^b^((ln3)/(ln5))    (49^a )^(((ln5)/(ln3))×((ln3)/(ln5)))   49^a =p  ln_(49) p=a  ((lnp)/(2ln7))=a=((ln6)/(ln7))  lnp=2ln6=ln6^2   p=36
$$ \\ $$$$\left(\mathrm{49}^{{log}_{\mathrm{7}} ^{\mathrm{6}} } \right)^{{log}_{\mathrm{3}} \mathrm{5}^{{log}_{\mathrm{2}\:} \mathrm{3}^{{log}_{\mathrm{5}\:} \mathrm{2}} } } \\ $$$$\left(\mathrm{49}^{{a}} \right)^{{b}^{\frac{{ln}\mathrm{2}}{{ln}\mathrm{5}}×\frac{{ln}\mathrm{3}}{{ln}\mathrm{2}}} } \\ $$$$\left(\mathrm{49}^{{a}} \right)^{{b}^{\frac{{ln}\mathrm{3}}{{ln}\mathrm{5}}} } \\ $$$$\left(\mathrm{49}^{{a}} \right)^{\frac{{ln}\mathrm{5}}{{ln}\mathrm{3}}×\frac{{ln}\mathrm{3}}{{ln}\mathrm{5}}} \\ $$$$\mathrm{49}^{{a}} ={p} \\ $$$${ln}_{\mathrm{49}} {p}={a} \\ $$$$\frac{{lnp}}{\mathrm{2}{ln}\mathrm{7}}={a}=\frac{{ln}\mathrm{6}}{{ln}\mathrm{7}} \\ $$$${lnp}=\mathrm{2}{ln}\mathrm{6}={ln}\mathrm{6}^{\mathrm{2}} \\ $$$${p}=\mathrm{36} \\ $$

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