SOLUTION: Need help please! I am failing this class and just can't seem to grasp it.. Solve. Where appropriate, include approximations to the nearest thousandth. If no solution exists, stat

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Question 328048: Need help please! I am failing this class and just can't seem to grasp it..
Solve. Where appropriate, include approximations to the nearest thousandth. If no solution exists, state this.
log9x = -3
9 is base( i believe)
Answer by josmiceli(9686) About Me  (Show Source):
You can put this solution on YOUR website!
You have to be able to read this. First you must
know that **logs are exponents**.
If you have a%5Eb, then b is a log to the base a
log%289%2Cx%29+=+-3 says the log to the base 9 which gives
me x is -3.
You know -3 is the exponent because
log(something) = -3
If -3 is the exponent, and 9 is the base, I can write
9%5E%28-3%29+=+x
Now you have to know what to do with a negative exponent
The rule is:
a%5E%28-b%29+=+1%2Fa%5Eb, so
9%5E%28-3%29+=+1%2F9%5E3, and
1%2F9%5E3+=1%2F+%289%2A9%2A9%29
1%2F%289%2A9%2A9%29+=+1%2F729
So far, I have
9%5E%28-3%29+=++1%2F729, so
x+=+1%2F729
Remember, also, that there is no way you could have
something like
log%2810%2C-10%29+=+x
This is saying 10%5Ex+=+-10
There is no possible x that can give you a minus result
Good luck - just pound away at this stuff- It's like
cement- It will crumble after a while