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vivalakoda

  • 3 years ago

Find the solution to the equation 64^(4 – x) = 4^(2x)?

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  1. ParthKohli
    • 3 years ago
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    Notice:\[64 ^{4 - x} = (4^3)^{4 - x}\]

  2. vivalakoda
    • 3 years ago
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    Oh, 4^3 is equal to 64. So would I make it: \[64^{4-x}=64^{4-x}\]

  3. lacypennelll
    • 3 years ago
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    :\ So look in your text book yet?

  4. vivalakoda
    • 3 years ago
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    Sadly I don't have a textbook, I do online school. I've been trying to reach my teacher all morning but she hasn't responded :/

  5. lacypennelll
    • 3 years ago
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    I do online school to I know how it feels when a teacher doesn't reach you back =-=

  6. anonymous
    • 3 years ago
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    do what @ParthKohli said rewrite \(64\) as \(4^3\) this means \(64^{4-x}=4^{3(4-x)}=4^{12-3x}\)

  7. anonymous
    • 3 years ago
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    then you know \(12-3x=2x\) so you can solve for \(x\)

  8. precal
    • 3 years ago
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    do what satellite73 tells you to do if you can create the same bases then you can set the exponents equal to each other and solve for x

  9. vivalakoda
    • 3 years ago
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    Thank you everyone for the help, I understand the problem now.

  10. precal
    • 3 years ago
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    yw

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