SOLUTION: How do I solve this problem? What am I supposed to do with the numbers in parenthesis? Multiply them? log b (x+2) + log b (x-4)= log b 24 its similar to this one which I am

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: How do I solve this problem? What am I supposed to do with the numbers in parenthesis? Multiply them? log b (x+2) + log b (x-4)= log b 24 its similar to this one which I am       Log On


   

Question 649010: How do I solve this problem? What am I supposed to do with the numbers in parenthesis? Multiply them?
log b (x+2) + log b (x-4)= log b 24
its similar to this one which I am also having trouble with:
log10 (x+6) - log10 (x-3)=1
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
log b (x+2) + log b (x-4)= log b 24
its similar to this one which I am also having trouble with:
log10 (x+6) - log10 (x-3)=1
**
log b (x+2) + log b (x-4)= log b 24
logb[(x+2)(x-4)]=logb(24)
(x+2)(x-4)=24
x^2-2x-8=24
x^2-2x-32=0
solve for x by following quadratic formula:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
a=1, b=-2, c=-32
ans:
x≈-4.744 ( reject, (x+2)>0)
or
x≈6.744
note: base does not enter into calculations.
..
log10 (x+6) - log10 (x-3)=1
place under single log
log10[(x+6)/(x-3)]=1og10(10)
[(x+6)/(x-3)=10
x+6=10x-30
9x=36
x=4