SOLUTION: 1.) log[10](a)+log[10](a+21)=2 2.) log[6](a^2+2)+log[6](2)=2

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: 1.) log[10](a)+log[10](a+21)=2 2.) log[6](a^2+2)+log[6](2)=2      Log On


   

Question 295377: 1.)
log[10](a)+log[10](a+21)=2
2.)
log[6](a^2+2)+log[6](2)=2
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
1) Solve for a:
Log%5B10%5D%28a%29%2BLog%5B10%5D%28a%2B21%29+=+2 Apply the product rule for logarithms:Log%5Bb%5D%28M%29%2BLog%5Bb%5D%28N%29+=+Log%5Bb%5D%28M%2AN%29
Log%5B10%5D%28a%29%2BLog%5B10%5D%28a%2B21%29+=+Log%5B10%5D%28a%2A%28a%2B21%29%29 Simplify the argument on the right side.
Log%5B10%5D%28a%5E2%2B21a%29+=+2 Rewrite this in exponential form.
10%5E2+=+a%5E2%2B21a Rearrange into the standard quadratic equation form.
a%5E2%2B21a-100+=+0 Solve by factoring.
%28a-4%29%28a%2B25%29+=+0 Apply the zero product rule.
a-4+=+0 or a%2B25+=+0 so then...
a+=+4 or a+=+-25 Discard the negative solution*.
highlight%28a+=+4%29
Check: Substitute a = 4.
Log%5B10%5D%28a%29%2BLog%5B10%5D%28a%2B21%29+=+Log%5B10%5D%284%29%2BLog%5B10%5D%2825%29 = 2
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2) Solve for a.
Log%5B6%5D%28a%5E2%2B2%29%2BLog%5B6%5D%282%29+=+2 Apply the product rule.
Log%5B6%5D%282%2A%28a%5E2%2B2%29%29+=+2 Rewrite this in exponential form.
6%5E2+=+2a%5E2%2B4 Subtract 4 from both sides.
32+=+2a%5E2 Divide both sides by 2.
a%5E2+=+16 Take the square root of both sides.
a+=+4 or a+=+-4 Discard the negative solution*.
highlight%28a+=+4%29
*The logarithm of a negative quantity yields a complex number.