How do you solve #Log x + log(x-9)=1#?

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Answer 1 · 2015-08-10T00:15:08

#color(red)(x=10)#

# logx+log(x−9)=1#

Recall that #loga+logb=log(ab)#, so

# logx+log(x−9)=log(x(x-9))=log(x^2-9x)#

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

Convert the logarithmic equation to an exponential equation.

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

Remember that #10^logx =x#, so

#x^2-9x=10#

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

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

#x-10=0# and #x+1=0#

#x=10# and #x=-1#

Check:

# logx+log(x−9)=1#

If #x=10#,

# log10+log(10−9)=1#

#log10+log1=1#

#1+0=1#

#1=1#

#x=10# is a solution.

If #x=-1#,

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

This is impossible, because the logarithm of a negative number is undefined.