SOLUTION: How do you solve these algebraically: 1) log(x+1)+log(x-1)=log(3) 2) 5e^x+2=20 3) 4e^x=48 4) 9.5e^0.005x=19 5) 5log(base 3)x=10

Algebra ->  Equations -> SOLUTION: How do you solve these algebraically: 1) log(x+1)+log(x-1)=log(3) 2) 5e^x+2=20 3) 4e^x=48 4) 9.5e^0.005x=19 5) 5log(base 3)x=10       Log On


   

Question 169148: How do you solve these algebraically:
1) log(x+1)+log(x-1)=log(3)


2) 5e^x+2=20


3) 4e^x=48

4) 9.5e^0.005x=19

5) 5log(base 3)x=10


Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Solve algebraically:
1) Log%28x%2B1%29%2BLog%28x-1%29+=+Log%283%29 Apply the "product rule" (Log%5Bb%5D%28M%29%2BLog%5Bb%5D%28N%29+=+Log%5Bb%5D%28M%2AN%29) for logarithms to the left side.
Log%28%28x%2B1%29%28x-1%29%29+=+Log%283%29 Simplify the left side.
Log%28x%5E2-1%29+=+Log%283%29 Apply the identity: IfLog%5Bb%5D%28M%29+=+Log%5Bb%5D%28N%29, then M+=+N, so...
x%5E2-1+=+3 Add 1 to both sides.
x%5E2+=+4 Take the square root of both sides.
highlight%28x+=+2%29 or x+=+-2 Discard the negative solution as the log of a negative number is not a real number.
2) 5e%5E%28x%2B2%29+=+20 Divide both sides by 5.
e%5Ex+=+4 Take the natural log of both sides.
ln%28e%5Ex%29+=+ln%284%29 Apply the "power rule" for logs: Log%5Bb%5D%28M%5Ea%29+=+a%2ALog%5Bb%5D%28M%29
x%2Aln%28e%29+=+ln%284%29 Substitute ln%28e%29+=+1
x+=+ln%284%29
x+=+1.386
3) 4e%5Ex+=+48 Divide both sides by 4.
e%5Ex+=+12 Take the natural log of both sides.
x%2Aln%28e%29+=+ln%2812%29
x+=+ln%2812%29
highlight%28x+=+2.485%29
4) 9.5e%5E%280.005x%29+=+19 Divide both sides by 9.5
e%5E%280.005x%29+=+2 Take the natural log of both sides.
0.005%2Aln%28e%5Ex%29+=+ln%282%29 Divide both sides by 0.005
x+=+ln%282%29
highlight%28x+=+0.693%29
5) 5%2ALog%5B3%5D%28x%29+=+10 Divide both sides by 5.
Log%28x%29+=+2 Rewrite in exponential form: Log%5Bb%5D%28x%29+=+y means b%5Ey+=+x so...
3%5E2+=+x
highlight%28x+=+9%29