You can put this solution on YOUR website! The simplest way to solve this is to start by combining the two logarithms into one. This can be done using the following property of logarithms (which applies to any base):
log(x) + log(y) = log(x*y)
Using this on your problem:
logbase3(x+6) + logbase3(x) = 3
becomes
logbase3((x+6)*x) = 3
Now we can rewrite this in exponential form:
Simplifying both sides we get:
This is a quadratic equation we can solve by factoring. First get one side equal to zero by subtracting 27 from both sides giving:
Factoring we get:
For this product to be zero one of the factors must be zero. In other "words":
(x + 9) = 0 or (x - 3) = 0
Solving these separately we get
x = -9 or x = 3
Normally we would think we're done. But we cannot allow the argument to any logarithm be zero or negative. So we need to check our answers.
When we check x = -9, we find that the argument of logbase3(x+6) would be negative if x = -9 so we have to reject that solution. (It also makes the argument of logbase3(x) negative but that is kind of irrelevant at this point. We already had to reject -9 because of logbase3(x+6). It didn't make any difference whether or not logbase3(x) was negative.)
When we check x = 3 we find that the arguments to both logarithms, logbas3(x+6) and logbase3(x), work out to be positive. So x=3 is the one and only solution to your problem.
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