SOLUTION: Log(10+9x)-log(11-x)=2

Question 1166083: Log(10+9x)-log(11-x)=2
Found 2 solutions by Theo, htmentor:
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
your equation is log(10 + 9x) - log(11 - x) = 2
since log(a) - log(b) is equal to log(a/b), you get:
log( (10+9x)/(11-x) ) = 2
since log(a) = b if and only if 10^b = a, you get:
10^2 = (10+9x)/(11-x)
simplify 10^2 and multiply both sides of the equaion by (11-x) to get:
100 * (11-x) = 10+9x
simplify to get 1100 - 100x = 10 + 9x
subtract 10 from both sides of this equation and add 100x to boh sides of this equation to get:
1090 = 109x
solve for x to get:
x = 1090/109 = 10

your solution is x = 10

when x = 10, log(10 + 9x) - log(11 - x) = 2 becomes:
log(10 + 90) = log(11 - 10) which becomes:
log(100) - log(1) = 2
log(100) is equal to 2 and log(1) is equal to 0, so your equation becomes:
2 - 0 = 2 which becomes 2 = 2, confirming your solution is good.

Answer by htmentor(1343) (Show Source): You can put this solution on YOUR website!
I'm assuming log here means log base 10
Using the rule log(a) - log(b) = log(a/b) we have:
log((10+9x)/(11-x)) = 2
To eliminate the logarithm, we use the exponential form:
10^(log((10+9x)/(11-x))) = 10^2
(10+9x)/(11-x) = 100
Solving for x gives 1100 - 100x = 10 + 9x -> 109x = 1090 -> x = 10
Check: log(10+90) - log(1) = 2
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