Problem: To the nearest thousandth, log10​2 is .301 and log10​3 is .477. Which of the following is the best approximation of log5​10?
Answer Choices:
A. 78​
B. 79​
C. 710​
D. 711​
E. 712​
Solution:
Since logb​a=loga​b1​,log5​10=log10​51​ =log10​10−log10​21​≈.6991​≈710​. (The value of log10​3 was not used.)