SOLUTION: log x cubed root of 9 =1/6

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Question 398987: log x cubed root of 9 =1/6
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
When posting logarithmic equations, please indicate the base of the logarithm more clearly. For example, your equation could be posted as
base x log of cube root of 9 = 1/6
log base x of cube root of 9 = 1/6
or
log(x, (root(3, 9))) = 1/6
(If you put three left braces, "{" in front and three right braces, "}" behind the last one, your equation would look like it should:
log%28x%2C+%28root%283%2C+9%29%29%29+=+1%2F6
Tutors are more likely to help when a problem is clearly stated.

Solving an equation like this starts with rewriting the equation in exponential form. In general log%28a%2C+%28p%29%29+=+q is equivalent to p+=+a%5Eq. Using this pattern on your equation we get:
root%283%2C+9%29+=+x%5E%28%281%2F6%29%29
Now we can solve for x. The solution will be of the form
x = something
Since x+=+x%5E1 we can think of this as
x%5E1 = something
So we have x%5E%281%2F6%29 and we want x%5E1. Somehow we need to find a way to change the exponent from 1/6 to 1. To do this we will put together three ideas:
  • It is OK to raise both sides of an equation to the same power.
  • When raising a power to a power, the rule for exponents is to multiply the exponents.
  • When multiplying reciprocals, the result is always a 1!

So we can change the exponent from 1/6 to 1 by raising both sides of the equation to the reciprocal of 1/6 power. The reciprocal of 1/6 is 6/1 or 6:
%28root%283%2C+9%29%29%5E6+=+%28x%5E%28%281%2F6%29%29%29%5E6
On the right side we get x%5E1 (or x) as planned:
%28root%283%2C+9%29%29%5E6+=+x
Now we just have to simplify the left side. If we rewrite the radical with a fractional exponent this becomes pretty easy. root%283%2C+9%29+=+9%5E%281%2F3%29:
%289%5E%281%2F3%29%29%5E6+=+x
The rule for exponents when raising a power to a power is to multiply the exponents. So this becomes:
9%5E%28%281%2F3%29%2A6%29+=+x
which simplifies as follows:
9%5E2+=+x
81 = x
When solving logarithmic equations you must check your answers. The bases and arguments of all logarithms must be positive for any proper solution. Any solution that makes a base or an argument zero or negative must be rejected. These rejected solutions can occur even when no mistakes have been made. This is why you must check.

Always use the original equation to check:
log%28x%2C+%28root%283%2C+9%29%29%29+=+1%2F6
Checking x = 81:
log%28%2881%29%2C+%28root%283%2C+9%29%29%29+=+1%2F6
We can see that both the base and the argument are positive. 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 any mistakes. You are welcome to finish the check.

So x = 81 is the solution to your equation.