Answer:
log_2(100) is about 4 times the value of log_6(20)
Step-by-step explanation:
Lets find the value of both logs and then compare
\(log_2(100)\)
To find the value we use change of base formula
\(log_b(a) = \frac{log(a)}{log(b)}\)
\(log_2(100) = \frac{log(100)}{log(2)}= 6.643856\)
Now we find the value of log_6(20)
\(log_6(20) = \frac{log(20)}{log(6)}=1.67195\)
6.643556 is approximately four times of 1.67195
So log_2(100) is about 4 times the value of log_6(20)
