How do you solve #\log _ { 5} x = \frac { 1+ \log _ { 5} 125x } { 5}#?

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Answer 1 · 2017-09-25T12:08:42

Solution: #x=5#

#log_5x = (1+log_5 125x)/5 or log_5x = (1+log_5 125+log_5x)/5 or#

[Since #log_b (MN) = log_b M+log_b N#]

#5*log_5x = 1+log_5 125+log_5x) # or

#5*log_5x - log_5x= 1+log_5 5^3# or

#4*log_5x = 1+3#, [since #log_b b^n=n#] or

#cancel4*log_5x = cancel4 or log_5x = 1 :. x = 5^1=5#,

[If , #log_bN= x # then #b^x=N#]

Solution: #x=5# [Ans]