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How do you evaluate #3 log_3 (9) - 4 log_3 (3)#?

1 Answer

A. S. Adikesavan

May 19, 2016

2

Explanation:

Use #log _b a = log a/log b, n log a = log a^n and log_b b=1#.

Here, #3 log_3 9 - 4 log_3 3#

#=3(log 9/log 3)-4(1)#

#=3(log (3^2)/log 3)-4#

#=6(log 3/log 3)-4=6-4 = 2#

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