You can put this solution on YOUR website!
Solving equations like this, where the variable is in the argument of a logarithm usually starts with transforming the equation into one of the following forms:
log(expression) = other-expression
or
log(expression) = log(other-expression)
With the "non-log" term of 5 the second form will be more difficult to achieve. So we will aim for the first form.
If you really understand exponents and logarithms well, you would know that . Then you could just subtract 2/3 from each side and you would have the first form. But we are going to go ahead as if the second log was one that could not be found "by hand".
To achieve the first form, we need to find a way to combine the two logarithms into one. They are not like terms so we cannot just add them together. (Like logarithmic terms have the same bases and same arguments.)
But there is are properties of logarithms:
which provide other ways to combine logarithms. As long as the bases are the same and the coefficients are 1's we can use these properties. Your logarithms meet these requirements so we can use these properties. Your logarithms have a "+" between them so we will use the first property (which also has a "+" between the logs):
which simplifies to:
We now have 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 your equation we get:
We now have an "non-log" equation we can solve. It is a quadratic so we want one side to be zero. (Note: I am leaving unsimplified because I can see we will not need to do so. If this bothers you, you can go ahead and raise 8 to the 5th power.) Subtracting from each side we get:
Now we factor (or use the Quadratic Fomula). We can factor out an 8 to start with:
Now we can factor the difference of squares:
Now that the powers of 8 are easier, I'm going to simplify them:
From the Zero Product Property we know that one of these factors must be zero. The 8 can't be zero but the other two can:
x+64 = 0 or x-64 = 0
Solving these we get:
x = -64 or x = 64
When solving equations like yours, with the variable in the argument of a logarithm, you must check your answer(s). You must ensure that your solutions make all the arguments of all the logarithms positive. Any "solution" that makes an argument zero or negative must be rejected. These "solutions" that make an argument zero or negative can happen even if no mistakes were made. So you must always check.
Use the original equation to check:
Checking x = -64:
Since the -64 is being squared, we can already tell that the argument of the first log will be positive. And the argument of the second log is 4 which is also positive. So there is no reason to reject this solution. This is the required part of the check. The rest of the check is optional and will tell us if we made a mistake. You are welcome to finish the check if you like.
Checking x = 64:
Again we can already see that both arguments will be positive. So there is no reason to reject this solution, either. You can finish the check if you like.
So your equation has two solutions:
x = -64 or x = 64
Raise each side to the exponent with base 8, i.e. to get
x = 64 or -64. We have to be especially careful dealing with negative solutions or obtaining a negative in the logarithm, since the logarithm is not defined on negative numbers (disregarding complex values). We can check that both solutions satisfy.
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