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Question Number 122286 by ZiYangLee last updated on 15/Nov/20

$${\left({49}^{{\log }_{7}6}\right)}^{{\log }_3{5}^{{\log }_2{3}^{{\log }_{5}2}}}=?$$

Answered by TANMAY PANACEA last updated on 15/Nov/20

$$$$$${\left({49}^{{log}_7^6}\right)}^{{log}_3{5}^{{log}_2{3}^{{log}_{5}2}}}$$$${\left({49}^a\right)}^{b^{\frac{ln2}{ln5}\times \frac{ln3}{ln2}}}$$$${\left({49}^a\right)}^{b^{\frac{ln3}{ln5}}}$$$${\left({49}^a\right)}^{\frac{ln5}{ln3}\times \frac{ln3}{ln5}}$$$${49}^a=p$$$${ln}_{49}p=a$$$$\frac{lnp}{2ln7}=a=\frac{ln6}{ln7}$$$$lnp=2ln6=ln{6}^2$$$$p=36$$

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