How do you find the exact value of #log_8 32+log_8 2#?

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Answer 1 · 2017-12-11T16:24:13

# 2#.

Recall that, #log_bm+log_bn=log_b(mn).#

#:. log_8 32+log_8 2,#

#=log_8 (32xx2),#

#=log_8 64,#

#=log_8 8^2,#

#=2log_8 8,........[because, log_b(m^n)=nlog_bm],#

#=2xx1.................[because, log_b b=1].#

# rArr log_8 32+log_8 2=2#.