SOLUTION: solve for x: log x + (log x + 21)=2

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Question 382936: solve for x: log x + (log x + 21)=2
Found 2 solutions by Earlsdon, CharlesG2:
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for x:
Log%28x%29%2BLog%28x%2B21%29+=+2 Apply the product rule:
Log%28%28x%29%28x%2B21%29%29+=+2 Simplify the argument.
Log%28x%5E2%2B21x%29+=+2 Rewrite this in exponential form.
10%5E2+=+x%5E2%2B21x Rewrite as a quadratic equation in standard form.
x%5E2%2B21x-100+=+0 Factor.
%28x-4%29%28x%2B25%29+=+0 Apply the zero product rule.
%28x-4%29+=+0 or x%2B25+=+0 so...
x+=+4 or x+=+-25 Discard the negative solution as the log of a negative number is not real.
x+=+4
Check:
Log%284%29%2BLog%284%2B21%29+=+Log%284%29%2BLog%2825%29=2

Answer by CharlesG2(834) About Me  (Show Source):
You can put this solution on YOUR website!
solve for x: log x + (log x + 21)=2

log x + (log x + 21) = 2
thinking you meant log x + 21 and not log (x + 21), sorry if this was not the case
2log x + 21 = 2
2log x = -19
log x = -19/2 = -9.5
logarithmic rule: if logb x = y, then b^y = x
assuming base is 10
10^(-9.5) = x
3.162278 * 10^(-10) rounded places after the decimal point to 6 places