anonymous
  • anonymous
Given the function f(x) = 0.5(3)x, what is the value of f−1(7)?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
sO far, i have x=0.5(3)y
mathstudent55
  • mathstudent55
\(\Large f(x) = (0.5 ) 3^x\) \(\Large 7 = (0.5)3^x\) Solve for x.
anonymous
  • anonymous
7=1.5^x

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mathstudent55
  • mathstudent55
No. You can't multiply 0.5 by 3 since 3 is raised to x. What you can do is divide both sides by 0.5. Then take log of both sides.
mathstudent55
  • mathstudent55
Look at this example. See why you can't multiply the 0.5 and the 3 together because of the exponent. |dw:1439224190773:dw|
anonymous
  • anonymous
Oh! Yes, I understand noe :)
anonymous
  • anonymous
Going back to #1, I got 1.140 as answer.
mathstudent55
  • mathstudent55
\(\Large 7 = (0.5)3^x\) \(\Large \dfrac{7}{0.5} = \dfrac{(0.5)3^x}{0.5} \) \(\Large 14 = 3^x\) Now take log of both sides.
anonymous
  • anonymous
well log(14)=1.14612803568 and log(3)=0.47712125472
anonymous
  • anonymous
So then you divide??
mathstudent55
  • mathstudent55
I prefer to keep the answer exact, and not use decimal approximations. We take log of both sides: \(\Large \log 14 = \log 3^x\) \(\Large \log 14 = x \log 3\) Divide both sides by log 3: \(\Large \dfrac{\log 14}{\log 3} = x\) \(\Large x = \dfrac{\log 14}{\log 3} \)
anonymous
  • anonymous
x=2.402
mathstudent55
  • mathstudent55
That is fine if you are allowed to give an approximate answer. If you need an exact answer, then it's the fraction of logs above.
anonymous
  • anonymous
Ok:) Thanks! Can I ask one more?
mathstudent55
  • mathstudent55
Sure.
anonymous
  • anonymous
Given the parent functions f(x) = log2 (3x − 9) and g(x) = log2 (x − 3), what is f(x) − g(x)?
mathstudent55
  • mathstudent55
Subtract the functions. \(\Large f(x) = \log_2 (3x - 9) \) \(\Large g(x) = \log_2(x - 3)\) We subtract the second equation from the first equation: \(\Large f(x) - g(x) = \log_2(3x - 9) - \log_2(x - 3) \) Ok so far?
mathstudent55
  • mathstudent55
We're not done yet, but do you understand what we did so far?

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