SOLUTION: Solve the equation and put the solution in exact form. log(base 9)(x-4)+ log(base 9)(x-4)=1 This is what I have done..... log(base 9) (x-4)(x-4)=1 (x-4)(x-4)=9 x^2- 4x -4

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Solve the equation and put the solution in exact form. log(base 9)(x-4)+ log(base 9)(x-4)=1 This is what I have done..... log(base 9) (x-4)(x-4)=1 (x-4)(x-4)=9 x^2- 4x -4      Log On


   

Question 691275: Solve the equation and put the solution in exact form.
log(base 9)(x-4)+ log(base 9)(x-4)=1
This is what I have done.....
log(base 9) (x-4)(x-4)=1
(x-4)(x-4)=9
x^2- 4x -4x+16
x^2-8x+16=9
x^2-8x+7=0
(x-7)(x-1)
X=7, X=1
Answer (7,1)??? how would I put in exact form???? Is this even correct? Thanks!
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Your answers look right. Now you have to plug them
back into the original equation to verify them
+x+=+7+
I'll rewrite it slightly
+2%2Alog%28+9%2C+x-4+%29+=+1+
+2%2Alog%28+9%2C+7-4+%29+=+1+
+2%2Alog%28+9%2C+3+%29+=+1+
Note that +log%28+9%2C3+%29+=+1%2F2+
+2%2A%281%2F2%29+=+1+
OK
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+x+=+1+
Here you run into trouble. You get
+2%2Alog%28+9%2C+x-4+%29+=+1+
+2%2Alog%28+9%2C+1-4+%29+=+1+
+2%2Alog%28+9%2C+-3+%29+=+1+
It asks you for the log to the base 9 which
gives you a negative number. There is no real
number that can do this, so
+x+=+7+ is the only answer