SOLUTION: log(subscript 4)(2x+1)=log(subscript 4)(x-3)+log(subscript 4)(x+5) (so subscript is like pushing the downward arrow.)

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: log(subscript 4)(2x+1)=log(subscript 4)(x-3)+log(subscript 4)(x+5) (so subscript is like pushing the downward arrow.)       Log On


   

Question 205572: log(subscript 4)(2x+1)=log(subscript 4)(x-3)+log(subscript 4)(x+5)
(so subscript is like pushing the downward arrow.)
Found 2 solutions by Earlsdon, RAY100:
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for x:
Log%5B4%5D%282x%2B1%29+=+Log%5B4%5D%28x-3%29%2BLog%5B4%5D%28x%2B5%29 Apply the "product rule" for logarithms: Log%5Bb%5D%28M%29%2BLog%5Bb%5D%28N%29+=+Log%5Bb%5D%28MN%29 to the right side.
Log%5B4%5D%282x-1%29+=+Log%5B4%5D%28%28x-3%29%28x%2B5%29%29 Apply the identity rule:
If Log%5Bb%5D%28M%29+=+Log%5Bb%5D%28N%29 then M+=+N
%282x%2B1%29+=+%28x-3%29%28x%2B5%29 Expand the right side.
2x%2B1+=+x%5E2%2B2x-15 Subtract 1 from both sides..
2x+=+x%5E2%2B2x-16 Subtract 2x from both sides.
0+=+x%5E2-16 or...
x%5E2-16+=+0 Add 16 to both sides.
x%5E2+=+16 Take the square root of both sides.
highlight%28x+=+4%29 or highlight%28x+=+-4%29

Answer by RAY100(1637) About Me  (Show Source):
You can put this solution on YOUR website!
log(4) {2x+1} = log(4){x-3} + log(4){x+5}
.
log (2x+1) / log(4) = log(x-3)/log(4) + log(x+5) / log(4)
.
multiply thru by log(4)
.
log (2x+1) = log(x-3) + log (x+5)
.
log(2x+1) = log(x-3)(x+5),,,,,,loga +logb = log ab
.
2x+1 = (x-3)(x+5) = x^2 +2x -15
.
0 = x^2 -16,,,,factor
.
0 = (x+4)(x-4),,,,set equal to zero
.
x= 4, -4
.
check,,,(+4),,, log(4){9) = log(4){1)+log(4){9),,,but ,,,log(1)=0,,,,ok
.
(-4),,,log(4){-7} = log(4){-7}+log(4){1},,,but,,log(4){1}=0,,,,ok
.