help please Solve 64^x = 16^x−1. x = −2 x = −1 x = negative 1 over 4 x = negative 1 over 3

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help please Solve 64^x = 16^x−1. x = −2 x = −1 x = negative 1 over 4 x = negative 1 over 3

Mathematics
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HI!!
since \[4^2=16\] and \[4^3=64\] this is the same as \[\huge4^{3x}=4^{2(x-1)}\]
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Other answers:

now that the bases are the same, solve by solving \[3x=2(x-1)\] for \(x\)
I'm confused
me too
i sent a drawing of the problem
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
but the bases are not the same
ik
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
wanna go slow?
yes please
looking for cheap & free medals
ok your two bases are 64 and 16 right?
yes
and they are not equal
no they are not equal
but both 64 and 16 are powers of 4
because \[4^2=16\\ 4^3=64\] right?
yes that's correct
so...\[64^x=(4^3)^x=4^{3x}\] clear?
yes clear
how about \[16^{x-1}\] can you do the same thing with that one, like i did with \(64^x\)?
\[^{4^{2}}\]
well actually \[\huge (4^2)^{x-1}\]
yes that's correct
which is the same as \[\huge 4^{2(x-1)}\]
so now your equation looks like \[\huge 4^{3x}=4^{2(x-1)}\]and the bases are now the same
that means the exponents must also be the same, i.e. \[3x=2(x-1)\]
so would the answer be -1?
no i don't think so
can you solve \[3x=2(x-1)\]?

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