SOLUTION: I need help with: log7(x+4) +log7(x-2) = 1 <---thats to the base 7 not x7 I thought i had it but when checking the answer it wasn't right.

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: I need help with: log7(x+4) +log7(x-2) = 1 <---thats to the base 7 not x7 I thought i had it but when checking the answer it wasn't right.      Log On


   

Question 911885: I need help with:
log7(x+4) +log7(x-2) = 1 <---thats to the base 7 not x7
I thought i had it but when checking the answer it wasn't right.
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
+log%287%2C%28x%2B4%29%29+%2Blog%287%2C%28x-2%29%29+=+1+
+log%287%2C%28x%2B4%29%28x-2%29%29+=+1+ ...........log%287%2C7%29+=+1

+log%287%2C%28x%2B4%29%28x-2%29%29+=+log%287%2C7%29 same log =>
+%28x%2B4%29%28x-2%29+=+7
+x%5E2-2x%2B4x-8+=+7
+x%5E2%2B2x-8+-7=+0
+x%5E2%2B2x-15=+0+ .......2x as 5x-3x
x%5E2%2B5x-3x-15+=+0 ...group
%28x%5E2%2B5x%29-%283x%2B15%29+=+0 ...factor
x%28x%2B5%29-3%28x%2B5%29+=+0
%28x-3%29%28x%2B5%29+=+0
solutions:
if %28x-3%29=+0 => highlight%28x=3%29 ...we need only this one since log
if %28x%2B5%29=+0 => x=-5