How do you solve #log_10(x-3)^2=5.4#?

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Answer 1 · 2017-05-31T17:54:29

Given: #log_10(x-3)^2=5.4#

Bring the exponent outside of logarithm's argument by making it a coefficient:

#2log_10(x-3)=5.4#

Divide both sides by 2:

#log_10(x-3)=2.7#

Make both sides the exponent of 10:

#10^(log_10(x-3))=10^2.7#

The property of a function of its inverse makes the left side become #x - 3#:

#x - 3 = 10^(2.7)#

Add 3 to both sides:

#x =10^2.7+3#

#x ~~ 504.19#