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Solve the equation below. Please show all of your work. log2(1/64)=1/3-x Solve the equation below. Please show all of your work. log2(1/64)=1/3-x @Mathematics
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x= -log2(1/64) + 1/3 x= 19/3
how did you get this?
because loga +logb=log(a*b) so this is property of logaritms
but i think that this not is correct because so 1/64 = 1/2(6) like 2 on exponent 6 so log2(
so will be log2(2) on exponent minus 6
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than will be -6log2(2) = -6
than -6=1/3 -x so -18=1-3x -19=-3x 19=3x x=19/3
it's all numbers... log2(1/64) is a number (-6) just like 1/3 is a number (.3333) think of it as -6=.333-x and solve for x
log2(1/64) = log(1/64)/log(2)
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