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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(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 $b =$

Similar Questions

If $\log _{10} 7 = a$,then $\log _{10} \left( \frac{1}{70} \right)$ is equal to

If $\log (x-y) - \log 5 - \frac{1}{2} \log x - \frac{1}{2} \log y = 0$,then find the value of $\frac{x}{y} + \frac{y}{x}$.

If $a^{x}=b^{y}=c^{z}=d^{w},$ then $\log _{a}(b c d)=$

If $\log (2 a-3 b)=\log a-\log b,$ then $a=$

Evaluate: $2 \log _{3} 7 - 7 \log _{3} 2$

Vedclass Test Series

Exam Paper Generator

Online Exam Module