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miss_karina

  • 2 years ago

Find x when 2^x=16^y, 16^x=8^(y-1) Okay when I exponentiate for the first part I get y=x/3 and for the second part I get y=4x-1/3. 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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  1. satellite73
    • 2 years ago
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    \[2^x=16^y\] \[2^x=2^{4y}\] \[x=4y\]

  2. satellite73
    • 2 years ago
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    then \[16^x=8^{y-1}\] \[2^{4x}=2^{3(y-1)}\] \[4x=3y-3\]

  3. satellite73
    • 2 years ago
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    should be good from there yes?

  4. miss_karina
    • 2 years ago
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    okay 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.

  5. miss_karina
    • 2 years ago
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    There is only one answer and I am confused because I am given two equations for one problem

  6. khuangg
    • 2 years ago
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  7. miss_karina
    • 2 years ago
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    I have a question @khuangg , howcome on 16^x=8^(y-1) you distributed the exponent 3 in to (y-1) but on the other side you didn't distribute the 4 in to the x?

  8. miss_karina
    • 2 years ago
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    I still can't figure this out =(

  9. Azteck
    • 2 years ago
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    @khuangg is wrong.

  10. Azteck
    • 2 years ago
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    @miss_karina Have you learnt the topic Indices?

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