If log 2 = 0.3010 and log 3 = 0.4771, then th | vedclass

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

Question

(C) Given: $\log 2 = 0.3010$ and $\log 3 = 0.4771$.We need to find the value of $\log _{5} 512$.First,express $512$ as a power of $2$: $512 = 2^9$.Usi

Vedclass · Accepted Answer

$3.876$

Vedclass · Answer

(C) Given: $\log 2 = 0.3010$ and $\log 3 = 0.4771$.We need to find the value of $\log _{5} 512$.First,express $512$ as a power of $2$: $512 = 2^9$.Using the change of base formula,$\log _{b} a = \frac{\log _{10} a}{\log _{10} b}$.Therefore,$\log _{5} 512 = \log _{5} (2^9) = 9 \log _{5} 2$.This can be written as $9 	imes \frac{\log _{10} 2}{\log _{10} 5}$.Since $\log _{10} 5 = \log _{10} (10/2) = \log _{10} 10 - \log _{10} 2 = 1 - 0.3010 = 0.6990$.Substituting the values: $\log _{5} 512 = \frac{9 	imes 0.3010}{0.6990} = \frac{2.709}{0.699} = 3.8755... \approx 3.876$.

If $\log 2 = 0.3010$ and $\log 3 = 0.4771,$ then the value of $\log _{5} 512$ is

Similar Questions

If $\log 2=x, \log 3=y$ and $\log 7=z,$ then the value of $\log (4 \sqrt[3]{63})$ is

$\frac{3+\log _{10} 343}{2+\frac{1}{2} \log _{10} \left(\frac{49}{4}\right)+\frac{1}{3} \log _{10} \left(\frac{1}{125}\right)}=$

If $\log _{x} y=100$ and $\log _{2} x=10,$ then the value of $y$ is

If $\log _{8} p=2.5$ and $\log _{2} q=5$,then $p$ in terms of $q$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module