Question

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

Answer 1

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)

Answer 2

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!

Answer 3

\[\large 2^{x}=16^{x-1}\]
\[\large 2^{x}=\frac{16^{x}}{16}\]
\[\large 16=\frac{16^{x}}{2^{x}}=8^{x}\]
\[\large x \log8=\log16\]
\[\large x=\frac{\log16}{\log8}\]
\[\large x=\frac{4}{3}\]

Answer 4

The answer is actually b, which is 2. I understand what you did, but it's not correct?

Answer 5

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

Answer 6

Sure -

Answer 7

The person who wrote the official solution made an error.
See attachment and then ask questions about it if you like.