If log _{e} 2 cdot log _{b} 625 = log _{10} 1 | vedclass

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(B) Given equation: $\log _{e} 2 \cdot \log _{b} 625 = \log _{10} 16 \cdot \log _{e} 10$We know that $625 = 5^4$ and $16 = 2^4$. Substituting these va

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$5$

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(B) Given equation: $\log _{e} 2 \cdot \log _{b} 625 = \log _{10} 16 \cdot \log _{e} 10$We know that $625 = 5^4$ and $16 = 2^4$. Substituting these values:$\log _{e} 2 \cdot \log _{b} (5^4) = \log _{10} (2^4) \cdot \log _{e} 10$Using the property $\log(a^n) = n \log a$:$\log _{e} 2 \cdot 4 \log _{b} 5 = 4 \log _{10} 2 \cdot \log _{e} 10$Using the change of base formula $\log _{10} 2 = \frac{\log _{e} 2}{\log _{e} 10}$:$\log _{e} 2 \cdot 4 \log _{b} 5 = 4 \cdot \left( \frac{\log _{e} 2}{\log _{e} 10} ight) \cdot \log _{e} 10$Simplifying the right side:$\log _{e} 2 \cdot 4 \log _{b} 5 = 4 \log _{e} 2$Dividing both sides by $4 \log _{e} 2$ (assuming $\log _{e} 2 
eq 0$):$\log _{b} 5 = 1$By the definition of logarithm,$b^1 = 5$,so $b = 5$.

If $\log _{e} 2 \cdot \log _{b} 625 = \log _{10} 16 \cdot \log _{e} 10$,then $b =$

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