How do you solve #log_7x=log_2 9#?

1 Answer
Shwetank Mauria
Mar 25, 2016

#x=477.59#

Explanation:

As #log_b x=log_ax/log_ab#, we have

#log_7 x=log_2 9# is #log_10x/log_10 7=log_10 9/log_10 2# i.e.

#logx/log7=log9/log2#

Hence #logx=(log9*log7)/log2=(0.9542*0.8451)/0.3010# or

#logx=2.679# or #x=10^(2.679)# or

#x=477.59#

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