SOLUTION: Solve the following equation Log3x=log4(x+6)

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Question 623346: Solve the following equation
Log3x=log4(x+6)
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
I assume the problem is:
log%28%283x%29%29=log%28%284%28x%2B6%29%29%29
and not
log%283%2C+%28x%29%29=log%284%2C+%28x%2B6%29%29%29
If I am wrong, then please re-post your problem. Logarithms, especially logarithms of bases other than 10 (log) or e (ln), can be very tricky. Please use some English like
"base 3 log of x" and "base 4 log of (x+6)"
or learn the syntax used on algebra.com to display mathematical expressions. Click on the "Show Source" link above to see what I typed for these other logs.

If the problem is
log%28%283x%29%29=log%28%284%28x%2B6%29%29%29
then the solution is quite simple. The only way these two base 10 logs can be equal is if the arguments are equal. So
3x = 4(x+6)
This is easy to solve. Distribute the 4 on the right side:
3x = 4x+24
Subtract 4x from each side:
-x = 24
Divide by -1:
x = -24

When solving equations like this. where the variable is in the argument of a log, you must check your solution(s). It is not optional. You must ensure that all arguments are positive. Use the original equation to check:
log%28%283x%29%29=log%28%284%28x%2B6%29%29%29
Checking x = -24:
log%28%283%28-24%29%29%29=log%28%284%28%28-24%29%2B6%29%29%29
We can already see that both arguments are going to turn out negative. So we must reject this solution. Since it was the only one we found, it means that there are no solutions to:
log%28%283x%29%29=log%28%284%28x%2B6%29%29%29