anonymous
  • anonymous
help please Solve 64^x = 16^x−1. x = −2 x = −1 x = negative 1 over 4 x = negative 1 over 3
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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misty1212
  • misty1212
HI!!
misty1212
  • misty1212
since \[4^2=16\] and \[4^3=64\] this is the same as \[\huge4^{3x}=4^{2(x-1)}\]
anonymous
  • anonymous
|dw:1442080279656:dw|

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misty1212
  • misty1212
now that the bases are the same, solve by solving \[3x=2(x-1)\] for \(x\)
anonymous
  • anonymous
I'm confused
misty1212
  • misty1212
me too
anonymous
  • anonymous
i sent a drawing of the problem
misty1212
  • misty1212
the idea is this: if \[\huge b^{\spadesuit}=b^{\heartsuit}\] then \[\spadesuit =\heartsuit\] in other words, if the bases are the same, then so are the exponents
anonymous
  • anonymous
but the bases are not the same
misty1212
  • misty1212
ik
misty1212
  • misty1212
that is why i made them the same did you look at the answer i wrote above? i arranged it so the bases were equal
misty1212
  • misty1212
wanna go slow?
anonymous
  • anonymous
yes please
Mr_Perfection_xD
  • Mr_Perfection_xD
looking for cheap & free medals
misty1212
  • misty1212
ok your two bases are 64 and 16 right?
anonymous
  • anonymous
yes
misty1212
  • misty1212
and they are not equal
anonymous
  • anonymous
no they are not equal
misty1212
  • misty1212
but both 64 and 16 are powers of 4
misty1212
  • misty1212
because \[4^2=16\\ 4^3=64\] right?
anonymous
  • anonymous
yes that's correct
misty1212
  • misty1212
so...\[64^x=(4^3)^x=4^{3x}\] clear?
anonymous
  • anonymous
yes clear
misty1212
  • misty1212
how about \[16^{x-1}\] can you do the same thing with that one, like i did with \(64^x\)?
anonymous
  • anonymous
\[^{4^{2}}\]
misty1212
  • misty1212
well actually \[\huge (4^2)^{x-1}\]
anonymous
  • anonymous
yes that's correct
misty1212
  • misty1212
which is the same as \[\huge 4^{2(x-1)}\]
misty1212
  • misty1212
so now your equation looks like \[\huge 4^{3x}=4^{2(x-1)}\]and the bases are now the same
misty1212
  • misty1212
that means the exponents must also be the same, i.e. \[3x=2(x-1)\]
anonymous
  • anonymous
so would the answer be -1?
misty1212
  • misty1212
no i don't think so
misty1212
  • misty1212
can you solve \[3x=2(x-1)\]?

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