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Solving equations where the variable is in the argument of a logarithm usually starts with transforming the equation into one of the following forms:
log(expression) = number
or
log(expression) = log(expression)
If we can find a way to combine the two logs into one the equation will be in the second form. The two logs are not like terms so we cannot just subtract them. (Like logarithmic terms have the same bases and the same arguments. Your logs have the same base, e, but the arguments, x+5 and x-2, are different.) Fortunately there is a property of logarithms that allow us to combine two logs with a "-" between them: . This property requires the same bases and coefficients of 1. Your logs meet both requirements. Using this property we get:
We now have the equation in the first form.
The next step with this form is based on some simple logic. Logarithms are exponents. In our equation the log on the left is the exponent to put on a 2 to get (x+20)/(x+2) as a result. The log on the right is the exponent to put on a 2 to get x as a result. The equation says these two exponents are equal. Since putting the same exponent on a two will always result in the same result, the two arguments must be equal, So:
Now that the variable is out of the logarithms we can solve for it. We'll start by multiplying each side by x+2 to eliminate the fraction:
which simplifies to:
This is a quadratic equation. So we want one side to be zero. Subtracting the entire left side from each side we get:
Now we factor (or use the Quadratic Formula). This factors fairly easily:
From the Zero Product Property: or
Solving each of these we get: or
Last of all, we check our solutions. This is not optional! When solving equations where the variable is in the argument of a logarithm you must check that your "solutions" will make all arguments of all logs positive. Any "solution" that makes any argument zero or negative must be rejected because arguments of logarithms must be positive.
Use the original equation to check:
Checking x = -5:
We can already see that the last two arguments are negative if x = -5. So we must reject this solution.
Checking x = 4:
We can already see that all arguments are positive when x = 4. So this solution passes the required part of the check.
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