Question 401274
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.<br>
{{{log(3, (y)) = (1/4)log(3, (16)) + (1/3)log(3, (64))}}}
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)<br>
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.<br>
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 <i>and same argument</i>. Your logarithms have the same base but the arguments are different,)<br>
Although we cannot add the terms, there are properties of logarithms:<ul><li>{{{log(a, (p)) + log(a, (q)) = log(a, (p*q))}}}</li><li>{{{log(a, (p)) - log(a, (q)) = log(a, (p/q))}}}</li></ul>which allow us to combine two logarithms into one. Both of these properties require:<ul><li>Logarithms of the same base; and</li><li>A "+" between the two logarithms (for the first property) or a "-" between them (for the second property); and</li><li>Both logarithms have coefficients of 1.</li></ul>
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, {{{q*log(a, (p)) = log(a, (p^q))}}}, 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:
{{{log(3, (y)) = log(3, (16^(1/4))) + log(3, (64^(1/3)))}}}
We can now use the first property to combine these terms. (We use the first property because of the "+" between them.)
{{{log(3, (y)) = log(3, (16^(1/4)*64^(1/3)))}}}
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 {{{16 = 2^4}}} the 4th root of 16 is 2.  And since {{{64 = 4^3}}} the cube root of 64 is 4. Substituting these values in we get:
{{{log(3, (y)) = log(3, (2*4))}}}
which simplifies to
{{{log(3, (y)) = log(3, (8))}}}
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<br>
When solving logarithmic equations like this one, you <i>must</i> 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 <i>even if no mistakes have been made!</i> This is why it is not optional to check.<br>
Use the original equation to check:
{{{log(3, (y)) = (1/4)log(3, (16)) + (1/3)log(3, (64))}}}
Checking y = 8:
{{{log(3, ((8))) = (1/4)log(3, (16)) + (1/3)log(3, (64))}}}
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.