SOLUTION: Thank you so much for helping in this problem! And thank you for your time! Could you please guide me through this? Give an exact and decimal answer. 2^x = 9. Thank yo

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Thank you so much for helping in this problem! And thank you for your time! Could you please guide me through this? Give an exact and decimal answer. 2^x = 9. Thank yo      Log On


   

Question 635445: Thank you so much for helping in this problem! And thank you for your time!
Could you please guide me through this?
Give an exact and decimal answer.
2^x = 9.

Thank you!
Found 2 solutions by Tatiana_Stebko, jsmallt9:
Answer by Tatiana_Stebko(1539) About Me  (Show Source):
You can put this solution on YOUR website!
2%5Ex+=+9 Take a log-base 2 from both sides of the equation
log%282%2C2%5Ex%29=log%282%2C9%29
x=log%282%2C9%29
Use a scientific calculator to calculate log%282%2C9%29=ln9%2Fln2=3.1699
x=3.1699

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
2%5Ex=9
Solving for variables that are in an exponent usually involves using logarithms. Any base of logarithm may be used. But some choices are better than others:
  • Choosing a base for the logarithm that matches the base of the exponent will result in the simplest possible exact expression for the solution.
  • Choosing a base that yhour calculator "knows", base 10 (log) or base e (ln), will result in a less simple expression but one that can easily be turned into a decimal approximation.
I will use both choices so you can see the difference.

Matching the bases.
Since our exponent's base is 2, we'll use base 2 logs:
log%282%2C+%282%5Ex%29%29=log%282%2C+%289%29%29
Next we use a property of logarithms, log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29, which allows us to move the exponent of the argument out in front of the log. (It is this property that is the very reason we use logarithms. Moving the exponent, where the variable is, out in front we can now use "regular" algebra to solve for it. This property applies to all bases of logarithms which is why any base can be used.) Moving the exponent out in front we get:
x%2Alog%282%2C+%282%29%29=log%282%2C+%289%29%29
log%282%2C+%282%29%29+=+1 (This is why matching bases results in the simplest expression.) So this simplifies to:
x+=+log%282%2C+%289%29%29
This is an exact solution to your equation.

Using a base your calculator "knows".
The steps are mostly the same as above so I'll only comment on the differences:
ln%282%5Ex%29=ln%289%29
x%2Aln%282%29=ln%289%29
Here, the log on the left does not disappear like log%282%2C+%282%29%29 did. So we divide both sides by it:
x+=+ln%289%29%2Fln%282%29
This is another exact expression for the solution to your equation. It is not as simple as what we got earlier but it is easy to use your calculator to turn this into a decimal approximation. I'll leave that up to you with one caution: Be sure to find the two logs first and then divide. You can't divide 9 by 2 and then find the ln.