Question

How do you approximate `log_7 27` given `log_7 3=0.5646`?

Answer 1

1.694, nearly.

Explanation:

Just the value of `log_7 3` is sufficient,

for approximating `log_7 27`.

Use `log_b(a^n)=n log_b a`.

Here, #log_7 27 = log_7(3^3)=3 log_7 3= 3(.5646) = 1.694 ( rounded

to the limit of 4-sd )