SOLUTION: Need Help Please! I am failing this class.. 5. Solve. Where appropriate, include approximations to the nearest thousandth. If no solution exists, state this. logx (log3 9) =

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Need Help Please! I am failing this class.. 5. Solve. Where appropriate, include approximations to the nearest thousandth. If no solution exists, state this. logx (log3 9) =      Log On


   

Question 328052: Need Help Please!
I am failing this class..
5. Solve. Where appropriate, include approximations to the nearest thousandth. If no solution exists, state this.
logx (log3 9) = 2
Found 2 solutions by rapaljer, josmiceli:
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
First calculate log%283%2C9%29=y by converting to exponential notation: 3%5Ey=9, so y+=2

Now, you need to find log%28x%2Clog%283%2C9%29%29=2

You need to solve for x in this equation:
log%28x%2C2%29=2

Translate to exponential notation:
x%5E2=2, so x=+sqrt%282%29, which is approximately 1.414.

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Seminole State College of Florida
Altamonte Springs Campus

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Wow, this looks so confusing, I'm not sure what it says.
The log%283%2C9%29 is where to start, since it looks simple.
It says "I am an exponent to the base 3 that gives 9
I know that 3%5E2+=+9, so
log%283%2C9%29+=+2
Now I have
log%28x%2C2%29+=+2
This is saying "An exponent which is 2, with a base that is x
gives me 2
I'll rewrite it as:
x%5E2+=+2
Now take the square root of both sides
x+=+sqrt%282%29 answer
and
x+=+-sqrt%282%29 not answer
Only the positive result would work here, because what would the
answer to this be?
log%28-sqrt%282%29%2C+.3027%29
Can't solve it