How do you evaluate #\log _ { 7} 3+ \frac { \log _ { 7} 5} { 2} + \frac { \log _ { 7} 11} { 2}#?

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Answer 1 · 2017-05-20T09:11:38

#log_7 sqrt(495)#

#approx log_7 22.25#

#approx 1.59#

Before beginning, you must know the following properties related to logarithms:-

1. #a*log_cb = log_c(b^a)#
2. #log_ca + log_cb = log_c(a*b)#

Now, in this problem,

#log_7 3 + log_7 5/2 + log_7 11/2#

#=[2*log_7 3 + log_7 5 + log_7 11]/2#

#= [log_7 3^2 + log_7 (5*11)]/2#

#= [log_7 9 + log_7 55]/2#

#= [log_7 (9*55)]/2#

#=1/2*log_7 495#

#= log_7 (495^(1/2))#

#= log_7 sqrt(495)#

#approx log_7 22.25#

#approx 1.59#