Solving Logarithmic Equations
Solving logarithmic equations usually requires using properties of logarithms. The reason you usually need to apply these properties is so that you will have a single logarithmic expression on one or both sides of the equation.

Once you have used properties of logarithms to condense any log expressions in the equation, you can solve the problem by changing the logarithmic equation into an exponential equation and solving.

Let's Practice:
1. Solve
As mentioned above, the first step used in solving logarithmic equations is to make use of the properties of logs. In this case, we can combine the two log expressions on the left side of the equation into one expression using multiplication.
This means we are now solving the equation

Now that we have a single log expression equal to a number, we can change the equation into its exponential form. However, to do this we need to make note that when a base is not given in a problem, it is understood to be the common logarithm with a base of 10.

So our logarithmic equation becomes

.
Now we need to solve for x. This will require solving a quadratic equation by factoring. Note: Most of the time solving by factoring will suffice. Very seldom will you need to solve a quadratic by another method.

So let’s solve for x.

Factoring and setting each term equal to zero results in
(x - 5)(x + 2) = 0
x - 5 = 0    or     x + 2 = 0
x = 5    or     x = -2
When solving logarithmic equations, we must ALWAYS check our answers. Logarithmic functions are not defined for negative values. Therefore we have to plug in our answers and make sure we are not taking the log of a negative number.

We’ll plug in x = 5 into the original equation.
We do not actually have to continue in the checking process as soon as we see that we are not taking the log of a negative number.

Now let’s plug in x = -2 into the original equation.
As soon as we see that we are taking the log of a negative number we know that x = -2 is NOT a solution.
So our solution is x = 5.
1. Solve
This problem is slightly different than the last example we worked. First of all, it involves the natural logarithm (link to exponents-e.doc). But you may also notice that there are log expressions on both sides of the equation. Our approach to this type of problem is to write each side as a single log expression. Once we do that, we can apply the property of logarithms that says
If then x = y.
Let’s get started.

First we’ll apply properties of logs and write the left side of the equation as a single expression using multiplication and write the right side with an exponent of 2 rather than a coefficient of 2.
Now we use the property of logs shown above to get

Factoring and setting each term equal to zero results in

x - 6 = 0    or     x - 1 = 0
x = 6    or     x = 1
If we check x = 6 in the original equation we do not have to take the log of a negative number, but plugging x = 1 into the original equation will cause us to take the log of a negative number and cannot be a solution.
Our answer to this problem is x = 6.

Examples