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anonymous
 3 years ago
Find x when 2^x=16^y, 16^x=8^(y1)
Okay when I exponentiate for the first part I get y=x/3 and for the second part I get y=4x1/3. However, I don't understand how to get one x. I don't know what the next step is. Please help me.
anonymous
 3 years ago
Find x when 2^x=16^y, 16^x=8^(y1) Okay when I exponentiate for the first part I get y=x/3 and for the second part I get y=4x1/3. However, I don't understand how to get one x. I don't know what the next step is. Please help me.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[2^x=16^y\] \[2^x=2^{4y}\] \[x=4y\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then \[16^x=8^{y1}\] \[2^{4x}=2^{3(y1)}\] \[4x=3y3\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0should be good from there yes?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay I see how you got that but in order to find x do I have to set them equal to each other? I am given answers and it is not in that form. The answers are like in fraction form.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0There is only one answer and I am confused because I am given two equations for one problem

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I have a question @khuangg , howcome on 16^x=8^(y1) you distributed the exponent 3 in to (y1) but on the other side you didn't distribute the 4 in to the x?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I still can't figure this out =(

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@miss_karina Have you learnt the topic Indices?
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