You can put this solution on YOUR website!
Solving equations like this usually starts with using algebra and/or properties of logs to rewrite the equation in one of the following forms:
log(expression) = number
or
log(expression) = log(other_expression)
With the "non-log" term of 2 on the right side, it would seem that the "all-log" second form would be harder to achieve. So we will aim for the first form.
To achieve the first form all we need to do is find a way to combine the logs. Fortunately there is a property of logs, , which will do exactly what we need:
We now have the first form.
The next stage is to eliminate the logs. With the first form, you rewrite the equation in exponential form. In general is equivalent to . Using this pattern on our equation we get:
which simplifies to:
The next stage is to solve the equation (now that the logs are gone and the variables are exposed). First we'll eliminate the fraction by multiplying each side by x+3:
This is a quadratic equation so we want one side to be zero. Subtracting 9x and 27 from each side we get:
Now we factor:
(x-7)(x+2) = 0
Next the Zero Product Property:
x-7 = 0 or x+2 = 0
Solving these we get:
x = 7 or x = -2
The last stage is to check. This is not optional! A check must be made to ensure that the "solutions" make all the bases and arguments valid. (Valid bases are positive but not 1 and valid arguments are positive.) Any "solution" that makes any base or argument invalid must be rejected.
Use the original equation to check:
Checking x = 7:
We should be able to see already that the arguments are going to be positive and the bases are all 3's. (If you can't see this then keep simplifying.) So this solution checks.
Checking x = 7:
Simplifying...
And the arguments are positive (and the bases are still 3's) so this solution checks out, too. (Note: As this problem shows, negative numbers can be solutions. It is the arguments and bases which cannot be negative.
So there are two solutions to the equation: 7 and -2.
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