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Solving equations where the variable is in the argument (or base) of a logarithm usually starts with using algebra and/or properties of logarithms to transform the equation into on of the following forms:
log(expression) = other_expression
or
log(expression) = log(other_expression)
If we could combine the two logarithms on the left side of your equation we would have the first form. Fortunately there is a property of logarithms, , which allows to do just that:
whcih simplifies to:
We now have the first form.
With the first form, the next step with the first form is to rewrite the equation in exponential form. In general is equivalent to . Using this pattern on our equation we get:
which simplifies to:
We now have an equation without logs and where the variable is "out in the open". Now we use algebra to solve for x. Since this is a quadratic equation, we want one side of the equation to be a zero. Subtracting 10 from each side we get:
Now we factor (or use the quadratic formula):
(3x + 2)(x - 5) = 0
From the Zero Product Property we know that:
3x + 2 = 0 or x - 5 = 0
Solving these we get:
x = -2/3 or x = 5
Checking solutions for equations like this one is not optional. You must ensure that all arguments to all logarithms are positive. Use the original equation to check:
Checking x = -2/3:
We can already see that the first argument is negative when x = -2/3. So we must reject this solution.
Checking x = 5:
We can already see that both arguments are going to be positive when x = 5. Since this was the only solution we found that we did not have to reject, this is the only solution to your equation.
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