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anonymous

  • 5 years ago

256^4x=64^x+3

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  1. anonymous
    • 5 years ago
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    take log on both sides and solve

  2. anonymous
    • 5 years ago
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    do you know how to take log?

  3. anonymous
    • 5 years ago
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    Or you can simply make the base the same

  4. dumbcow
    • 5 years ago
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    just what mooolen said 256 = 4^4 64 = 4^3 ->4^16x = 4^(3x+9)

  5. anonymous
    • 5 years ago
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    wait, is the +3 on the exponent of 64 or ist it (64^x) +3? If it is the former, then do what moolen said.

  6. anonymous
    • 5 years ago
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    \[256^{4x}=64^x+3\]\[256^4x - 64^x = 3\]\[4^{4(4x)} - 4^{3x} =3\]\[4^{16x} - 4^{3x} =3\]\[4^{16x} - 4^{3x} =3 + 1 - 1\]\[4^{16x} - 4^{3x} =4^1 - 4^0\] since they all got same bases, you can simply equate the exponents. \[16x - 3x = 1-0\] \[x = 1/13\] NOT SURE IF I'VE DONE IT RIGHT. HOPE IT HELPED YOU.

  7. anonymous
    • 5 years ago
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    Sorry. I think I'm wrong

  8. amistre64
    • 5 years ago
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    256^(4x) = 64^(x+3) ???

  9. amistre64
    • 5 years ago
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    its hard to tell how the problem reads to begin with...

  10. anonymous
    • 5 years ago
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    i think the answer is x=0,9/13

  11. amistre64
    • 5 years ago
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    thats possible, but whats the question to begin with?

  12. amistre64
    • 5 years ago
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    256^(4x) = 64^(x+3) 4x log2(256) = (x+3) log2(64) 4x log2(64) ----- = -------- (x+3) log2(256) ---------------------- 4x = log2(64) x = log2(64)/4 -------------------- x+3 = log2(256) x = log2(256) -3

  13. amistre64
    • 5 years ago
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    if anything x = 9/13 :) if i see your question correctly

  14. amistre64
    • 5 years ago
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    if x = 0 the you get 1 = 64^3 which is not right

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