anonymous
  • anonymous
256^4x=64^x+3
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
take log on both sides and solve
anonymous
  • anonymous
do you know how to take log?
anonymous
  • anonymous
Or you can simply make the base the same

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More answers

dumbcow
  • dumbcow
just what mooolen said 256 = 4^4 64 = 4^3 ->4^16x = 4^(3x+9)
anonymous
  • anonymous
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.
anonymous
  • anonymous
\[256^{4x}=64^x+3\]\[256^4x - 64^x = 3\]\[4^{4(4x)} - 4^{3x} =3\]\[4^{16x} - 4^{3x} =3\]\[4^{16x} - 4^{3x} =3 + 1 - 1\]\[4^{16x} - 4^{3x} =4^1 - 4^0\] 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.
anonymous
  • anonymous
Sorry. I think I'm wrong
amistre64
  • amistre64
256^(4x) = 64^(x+3) ???
amistre64
  • amistre64
its hard to tell how the problem reads to begin with...
anonymous
  • anonymous
i think the answer is x=0,9/13
amistre64
  • amistre64
thats possible, but whats the question to begin with?
amistre64
  • amistre64
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
amistre64
  • amistre64
if anything x = 9/13 :) if i see your question correctly
amistre64
  • amistre64
if x = 0 the you get 1 = 64^3 which is not right

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