how does the value of $\\log_2 100$ compare with the value of $\\log_6 20$?\nthe value of $\\log_2 100$ is…

Explanation:

Step1: Use the change - of - base formula

The change - of - base formula is (\log_{a}b=\frac{\ln b}{\ln a}). For (\log_{2}100), by the change - of - base formula, (\log_{2}100 = \frac{\ln100}{\ln2}). Since (100 = 10^{2}), then (\log_{2}100=\frac{\ln(10^{2})}{\ln2}=\frac{2\ln10}{\ln2}\approx\frac{2\times2.3026}{0.6931}\approx6.64). For (\log_{6}20), by the change - of - base formula, (\log_{6}20=\frac{\ln20}{\ln6}). Since (20 = 2^{2}\times5) and (6 = 2\times3), then (\log_{6}20=\frac{\ln(2^{2}\times5)}{\ln(2\times3)}=\frac{2\ln2+\ln5}{\ln2+\ln3}\approx\frac{2\times0.6931 + 1.6094}{0.6931+1.0986}=\frac{1.3862 + 1.6094}{1.7917}=\frac{2.9956}{1.7917}\approx1.67).

Step2: Calculate the ratio

Calculate the ratio (\frac{\log_{2}100}{\log_{6}20}). (\frac{\log_{2}100}{\log_{6}20}\approx\frac{6.64}{1.67}\approx4).

Answer:

The value of (\log_{2}100) is about (4) times the value of (\log_{6}20).