can someone please look over this and tell me if I made a mistake?
log with a base of 3(−9) = 2x
-log with a base of 3(9) = 2x
-2=2x
x=-1
I'm pretty sure I did it right but just incase please look over and let me know if and where I went wrong. Thanks!
log(-9) is not defined
-log9 = log(1/9)
oh... ok thanks oobleck! Is that the only thing? Also what do you mean that log(-9) is not defined?
logs of negative numbers are not defined.
You want 3^x = -9
But all powers of 3 are positive. You have seen the graph, right?
To verify if you made a mistake, let's go through the steps of your solution:
1. The equation you started with is log with a base of 3(-9) = 2x.
2. To solve this equation, you can use the property of logarithms that states log base a of b = c is equivalent to a raised to the power of c equals b. Using this property, you can rewrite the equation as 3^(2x) = -9.
3. However, there is an issue with this step. It is not possible for a positive number (3 raised to any power) to equal a negative number. Therefore, there is no solution to the equation 3^(2x) = -9.
Hence, the mistake is in the third step where you concluded that x = -1.