You can put this solution on YOUR website!
Solving equations like this, where the variable is in the argument of a logarithm, usually starts by using algebra and/or properties of logs to transform the equation into one of the following forms:
log(expression) = other-expression
or
log(expression) = log(other-expression)
With the "non-log" term of 3 in your equation, it will be more difficult to rewrite it in the "all-log" second form. So we will aim for the first form. This means we will have to find a way to change the equation from having two logs into an equation with one log.
The two logs are not like terms. (Like logarithmic terms have the same bases and the same arguments.) So we cannot just subtract the terms.
Another way to combine log terms is to use one of the following properties of logs:
These properties require that the logs have the same bases and coefficients of 1. Your logs have the same bases but your coefficients are 2 and 3.
Fortunately, another property of logarithms, , can be used to move coefficients of logs into the argument as its exponent. So we start by using this property on both the logs in our equation:
which simplifies to:
Now that the coefficients are 1's we can use one of the other properties to combine the two logs into one. We use the second property because its logs, like yours, have a "-" between them:
which simplifies to:
We now have the first form. The next step with this form is to rewrite the equation in exponential form. In general is equivalent to . Using this pattern on our equation we get:
which simplifies to:
Now that the variable is out of the logs, we can use regular algebra to solve for it. Multiplying both sides by 27 we get:
Finding the 5th root of each side:
which simplifies as follows:
(