You can put this solution on YOUR website! I hope your equation is:
and not
or
If I am wrong then re-post your equation using parentheses around the arguments of logarithms and around fractions that are factors (like: ).
Solving equations like:
usually starts with transforming the equation into one of the following general forms:
log(expression) = number
or
log(expression) = log(expression)
Our equation is already in 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:
All we have left to do is simplify the left side. As your note indicates, 1/64 to the 3/2 would be:
What your note doesn't say is you can also use the cube of the square root:
Either can be used. So the question is: Which looks easier? Do you want to cube 1/64 and then find a square root? Or find the square root of 1/64 and then cube that? Since the square root of 1/64 is easy I like square root first and then cube:
When solving equations where the variable is in the argument (or base) of a logarithm, you should check your solutions to make sure the arguments (and bases) remain valid. (Valid arguments to logarithms must be positive. Valid bases for logarithms must be positive but not 1.) Use the original equation to check:
Checking x = 1/512:
Our argument and base are valid so our solution passes the required part of the check.
(