SOLUTION: Please help me solve x for logx3=4 Thank you in advance.

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Question 818106: Please help me solve x for logx3=4
Thank you in advance.
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Is the equation
log%28x%2C+%283%29%29+=+4
or
log%28%28x%5E3%29%29+=+4
or
log%28%28x%2A3%29%29+=+4?
In the future, please try to be clearer when posting logarithms. Try using more English. (For example log%28x%2C+%283%29%29+=+4 could be described as "the base x log of (3) = 4.) Or try teaching yourself algebra.com's formula syntax. Click on the "Show source" link above to see what I typed (and what you could type) to get the logs to display the way they do here.

Regardless of which equation is correct...
  1. Start by rewriting the equation in exponential form. Use the fact that log%28a%2C+%28p%29%29+=+n is equivalent to p+=+a%5En (and the fact "log" is a base 10 log).
  2. Solve the resulting equation.
  3. Then check your solution. (This is not optional!) Use the original equation to check and check that the bases and arguments are valid when x is the solution you found. (Reject any solution that makes a base or an argument invalid!)