Question 691267

Solve the equation and express the solution in exact form...

log9(x-4)+log9(x-4)=1


This is what I have done...
9=(x-4)(x-4)
9=(x^2-4x-4x+16)
-9           -9
0=(x^2-8x+7)
(x-7)(x-1)
x=7, x=1


not sure what Iam doing wrong.... Thanks!


{{{log (9, (x - 4)) + log (9, (x - 4)) = 1 }}}


{{{log (9, ((x - 4)(x - 4))) = 1}}}


{{{(x - 4)(x - 4) = 9^1}}}


{{{x^2 - 8x + 16 = 9}}}


{{{x^2 - 8x + 16 – 9 = 0}}}


{{{x^2 - 8x + 7 = 9}}}


(x - 1)(x - 7) = 0


x = 1 (ignore as this value will result in a NEGATIVE log value)


{{{highlight_green(x = 7 )}}}


Your answers are CORRECT. However, you need to do the check, which as you'll see will eliminate {{{x = 1}}} as a solution, thus leaving the only solution as {{{x = 7}}} (see above).


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