How do you solve #log_5 x + log_10 x = 4#?

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Answer 1 · 2016-03-25T07:21:09

#x=44.228#

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

#log_5x+log_10x=4# can be written as

#log_10x/log_10 5+log_10x=4# or

#log_10x/0.699+log_10x=4# or

#log_10x+0.699xxlog_10x=4xx0.699# or

#log_10x(1+0.699)=4xx0.699# or

#log_10x=(4xx0.699)/1.699=1.6457#

#x=10^(1.6457)=44.228#