How do you solve #log(x-3)=1-log(x)#?

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Answer 1 · 2015-12-06T23:32:20

#x=5#

Rearrange the expression to get

#log(x-3)+log(x)=1#

Now, use the property #log(a)+log(b)=log(a*b)#:

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

If by #log(x)# you mean the logarithm in base #10#, you can write #1# as #log(10)#, and so the expression becomes

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

Now use the fact that #log(a)=log(b) \iff a=b#:

#x(x-3)=10#

Expand:

#x^2-3x-10=0#

To solve this equation, we need two numbers #x_1# and #x_2# such that:

#x_1+x_2=3#
#x_1*x_2=-10#

These numbers are #5# and #-2#. We can't accept #-2# as a solution, because it would lead to the logarithm of a negative number.