Help meeee

vivia said:
2^(x+4)=p*2^(x)
Solve for p.
Please help meeeee!!!!
Thanks <3
Click to expand...
Hint:

a^(x+b) = (a^x) * (a^b) ......................................edited

Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this problem.
 
Professor Khan meant to write 2^(x+4) = 2^x*2^4 = 16*2^x

Please show us what you can do with that hint.
 
Hmm? Oh I found the answer by myself.
Isolate the variable by dividing the factors gradually.
p=16, am i right?
 
vivia said:
Hmm? Oh I found the answer by myself.
Isolate the variable by dividing the factors gradually.
p=16, am i right?
Click to expand...
I am not sure what you mean when you use the word gradually but the answer is 16. Good job!
 
For the sake of future students, I would do it this way.

[MATH]2^{(x + 4)} = p \cdot 2^x \implies log_2(2^{(x+4)}) = log_2( p \cdot 2^x) \implies [/MATH]
[MATH](x + 4) \cdot log_2(2) = log_2(p) + log_2(2^x) \implies x + 4 = log_2(p) + x \implies[/MATH]
[MATH]log_2(p) = 4 \implies p = 2^4 = 16.[/MATH]
I fully admit that logs are not need for this problem, which can be solved more easily. But logs are the basic tool for solving exponential equations that can solved exactly. So I would be consistent in application.
 
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