How do you solve #Log(x-9) = 3-Log(100x) #?

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Answer 1 · 2018-07-03T13:11:46

#color(blue)(x=10)#

#log(x-9)=3-log(100x)#

By the laws of logarithms:

#log(ab)=log(a)+log(b)color(white)(888)[1]#

#log(100x)=log(100)+log(x)#

Assuming these are base 10 logarithms:

#log(100)+log(x)=2+log(x)#

We now have:

#log(x-9)=3-2-log(x)#

#log(x-9)=1-log(x)#

Using #[1]#

#log(x-9)+log(x)=1#

#log(x(x-9))=1#

#log(x^2-9x)=1#

#10^(log(x^2-9x))=10^(1)#

#x^2-9x=10#

#x^2-9x-10=0#

Factor:

#(x+1)(x-10)=0=>x=-1 and x=10#

Checking solutions.

#x=-1#

#log((-1)-9)=3-log(100(-1))#

#log(-10)=3-log(-100)#

Logarithms are only defined for real numbers if for:

#log(x)#

#x>0#

Therefore #-1# is not a solution.

For #x=10#

#log(10-9)=3-log(100(10))#

#log(1)=3-log(1000)#

#0=3-3#

#0=0#

So #x=10# is the only solution.