anonymous
  • anonymous
Studying for the CLEP College Mathematics test and I found this question. I know the answer from a key, but I'm trying to understand how to solve it. Let 2^x = 16^(x-1). Which of the following is a solution? a) 1 + i and 1 - i b) 2 c) 1/2 + 3i and 1/2 - 3i d) 1 and -1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
I've been trying to solve by putting the solutions in for x, but I'm having some trouble dealing with the imaginary numbers. For instance, how to I solve 2^(1+i) = 16((1+i)-1)
anonymous
  • anonymous
If someone could just talk me through solving it I would really appreciate it. I am new to open study but I'll help you however I can!
kropot72
  • kropot72
\[\large 2^{x}=16^{x-1}\] \[\large 2^{x}=\frac{16^{x}}{16}\] \[\large 16=\frac{16^{x}}{2^{x}}=8^{x}\] \[\large x \log8=\log16\] \[\large x=\frac{\log16}{\log8}\] \[\large x=\frac{4}{3}\]

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anonymous
  • anonymous
The answer is actually b, which is 2. I understand what you did, but it's not correct?
Directrix
  • Directrix
The answer to the posted question is 4/3. I am thinking that the question or the options are both are incorrectly posted. To clear up the matter, would you post a screenshot of the actual question as it appears in your study materials. Thanks. @austenheroine29
anonymous
  • anonymous
Sure -
Directrix
  • Directrix
The person who wrote the official solution made an error. See attachment and then ask questions about it if you like.
Directrix
  • Directrix
|dw:1447533818856:dw|
Directrix
  • Directrix
Whoever worked out the official solution made a typo and wrote x^2 where just x was supposed to be. Then, the person made up options to go with the incorrectly worked problem. The whole thing is messed up. You might want to e-mail the company about that.
anonymous
  • anonymous
Thanks for you help! That makes a lot more sense.
Directrix
  • Directrix
Here is a substitute problem for you to work. Let me know what you get, okay.
Directrix
  • Directrix
Hint: Try to write both sides of the equation as powers of 2. 8 and 16 are both powers of 2.
Directrix
  • Directrix
@austenheroine29
anonymous
  • anonymous
Is the answer 6?
Directrix
  • Directrix
|dw:1447535484041:dw|
Directrix
  • Directrix
|dw:1447535649532:dw|
Directrix
  • Directrix
@austenheroine29
phi
  • phi
based on the answer, it looks like the typo is in the question, and they are asking \[ 2^{x^2} = 16^{x-1} \] as suggested above, try to get the "same base" on both sides using 16= 2^4 \[ 2^{x^2} = \left(2^4\right)^{x-1} \\ 2^{x^2} = 2^{4x-4} \] now equate exponents \[ x^2= 4x-4 \\ x^2 -4x+4= 0\] this factors into (x-2)(x-2)= 0 and x=2 (a repeated root)

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