How do you find the exact value of #log_5 75-log_5 3#?

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Answer 1 · 2017-01-07T05:42:52

#log_5 75 - log_5 3 = 2#

Remember that #log_a b - log_a c = log_a (b/c)#

#:. log_5 75 - log_5 3 = log_5 (75/3) = log_5 25#

Let #x = log_5 25#

#:. 5^x = 25#

#-> 5^x = 5^2#

Equating indices:

#x=2#

Hence: #log_5 75 - log_5 3 = 2#