# SOLVING LOGARITHMIC EQUATIONS

1. To solve a logarithmic equation, rewrite the equation in exponential form and solve for the variable.

Problem 5: Solve for x in the equation

Answer: is the exact answer and x-0.7639320225 is an approximate answer.

Solution:

Step 1: As we know by now, we can only take the logarithm of a positive number. Therefore, we need to make a restriction on the domain (values of x) so that the problem will be valid (have an answer).
The term is valid when x + 5 >0 or x > -5; the term is valid when x + 2 > 0 or x > -2; and the term is valid when x + 6 > 0 or x > -6. If we require that the domain be restricted to the set of all real numbers such that x > -2, all the terms will be valid.
Step 2: Simplify the left side of the original equation using Logarithmic Rule 1:

Step 3: We now have an equation of the form which implies that the expression a must equal the expression b or:

Step 4: You could also raise the base 2 to an exponent equal to the left side of the equation in Step 2, and you could raise the base 2 to an exponent equal to right side of the equation in Step 2:

When the base is the same as the base of the logarithm, the above equation can be simplified to

Step 5: Simplify the left side of the above equation:

Step 6: Subtract x and subtract 6 from both sides of the above equation:

Step 7: Use the quadratic formula to solve for x:

There are two exact answer: and and there are two approximate answers:

However only one of the answers is valid.
One of the answers ( ) is out of our specified domain of the set of all real numbers such that (x > -2).
Therefore, the exact and approximate answers are:

Check: Let's substitute the value in the original equation and determine whether the left side of the equation equals the right side of the equation after the substitution. In other words, does

or

or

or

or

The answer checks. We can also check our solution with the approximate answer. Does

or does

In another word,

Since the value of the left side of the original equation equals the right side of the original equation when we substitute the exact and the approximate value of x, We have proved the answer.
Let's illustrate why we had to throw out one of the answers. Let's check to see if the approximate solution (the one we discarded) works:

At this point we have to stop because we cannot take the log of a negative number. We simply cannot calculate the value if any of the terms are undefined.

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