How do you condense #log_5 (240) − log_5 (75) − log_5 (80)#?

1 Answer
Bdub
Mar 22, 2016

#log_5 240 - log_5 75 -log_5 80##=log _5(1/25)#->#=-2#

Explanation:

#log_5 240 - log_5 75 -log_5 80#
#=log_5(240/(75*80))#
#=log _5(1/25)#
#=log_5(1/5^2)#
#=log_5 5^-2#->Use property #1/x^n = x^-n#
#=-2* log_5 5#->use property #log_bx ^n=n*log_b x#
#=-2*1#->use property #log_b b=1#
#=-2#

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