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High School Mathematics
3.23 Problems on Logarithms - III

Example:
Show that, log (72/125) = 2 log 3 + 3 log 2 - 3 log 5.
Solution:
Taking factors of 72 and 125, we get
72 = 2 * 36
= 2 * 2 * 18
= 2 * 2 * 2 * 9
= 2 * 2 * 2 * 3 * 3
= 23 * 32
125 = 5 * 25
= 5 * 5 * 5
= 53
log (72/125) = log 72 - log 125. [Since log (a/b) = log a - log b]
= log (23 * 32) - log 53
= log 23 + log 32 - log 53 [Since log ab = log a + log b]
= 3 log 2 + 2 log 3 - 3 log 5. [Since log ab = b log a]
Hence prove the problem.
Directions: Prove the following. Also write at least ten examples of your own.

Show that, log (324/6125) = 4 log 3 + 2 log 2 - 3 log 5 - 2 log 7.

Show that, log (2000/567) = 4 log 2 + 3 log 5 - 4 log 3 - log 7.

Show that, log (250/1323) = log 2 + 3 log 5 - 3 log 3 - 2 log 7.

Show that, log (10125/847) = 3 log 5 + 4 log 3 - log 7 - 2 log 11.

Show that, log (1029/125) = log 3 + 3 log 7 - 3 log 5.

Show that, log (108/605) = 2 log 2 + 3 log 3 - log 5 - 2 log 11.