You can put this solution on YOUR website! I'm guessing that the 3's after "log" are the bases of the logarithms. You might want to make your logarithm problems clearer by using wording like:
base 3 log of ...
or
log base 3 of ...
Clearer problems usually get quicker responses from the tutors.
Solving equations where the variable is in the argument (or base) 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)
Since your equation has nothing but logarithmic terms the second form will probably be easier to achieve. All we need to do is to find a way somehow to combine the two logarithmic terms on the right into a single logarithmic term.
The two terms on the right are not like terms so we cannot just add them together. (Like terms with logarithms have the same base and same argument. Your logarithms have the same base but the arguments are different,)
Although we cannot add the terms, there are properties of logarithms:
which allow us to combine two logarithms into one. Both of these properties require:
Logarithms of the same base; and
A "+" between the two logarithms (for the first property) or a "-" between them (for the second property); and
Both logarithms have coefficients of 1.
Your two logarithms on the right side fit the first two requirements. But they have fractions for coefficients. not 1's. Fortunately there is another property of logarithms, , which allows one to move a coefficient of a logarithm into the argument as an exponent. So we will start by using this third property to move the coefficients:
We can now use the first property to combine these terms. (We use the first property because of the "+" between them.)
Before we proceed, let's simplify the argument on the right side. If you remember how fractional exponents work, you'll know that an exponent of 1/4 means 4th root and an exponent of 1/3 means cube root. Since the 4th root of 16 is 2. And since the cube root of 64 is 4. Substituting these values in we get:
which simplifies to
We now have the second form. With this form the next step is a bit of simple logic. The equation says that one base 3 logarithm is equal to another. In order for these logarithms of the same base to be equal, their arguments must be equal. So:
y = 8
When solving logarithmic equations like this one, you must check your solution(s)! One must make sure that all arguments (and bases) of all logarithms remain positive. A "solution" that causes an argument (or base) of a logarithm to become zero or negative must be rejected! A zero or negative argument (or base) can occur even if no mistakes have been made! This is why it is not optional to check.
Use the original equation to check:
Checking y = 8:
We can already see that all arguments and bases of every logarithm are positive when y = 8. So there is no reason to reject this solution. This is the required part of the check. The rest of the check will tell us if we made a mistake somewhere. You are welcome to finish the check.
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