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Sagot :
I'm going to assume that you've done logarithms ("logs") in class
before you ran into this one on homework.
3(w to the -2/3 power) + 6= 198
Subtract 6 from each side:
3(w to the -2/3 power) = 192
Take the log of each side:
log(3) -(2/3) log(w) = log(192)
Subtract log(3) from each side:
-(2/3) log(w) = log(192) - log(3) . . . notice that this side is log(192/3) = log(64)
-(2/3) log(w) = log(64)
Divide each side by -(2/3) :
log(w) = -(3/2) log(64)
Raise 10 to the power of each side:
w = (64) to the -3/2 power .
==> (A number) to the -3/2 power means 1/(the number to the +3/2 power).
==> The 3/2 power means either find the number's square root and then cube it,
or else find the number's cube and then square root it.
w = 1 / (64 to the 3/2 power) = 1 / 512 .
That's choice ' B '.
before you ran into this one on homework.
3(w to the -2/3 power) + 6= 198
Subtract 6 from each side:
3(w to the -2/3 power) = 192
Take the log of each side:
log(3) -(2/3) log(w) = log(192)
Subtract log(3) from each side:
-(2/3) log(w) = log(192) - log(3) . . . notice that this side is log(192/3) = log(64)
-(2/3) log(w) = log(64)
Divide each side by -(2/3) :
log(w) = -(3/2) log(64)
Raise 10 to the power of each side:
w = (64) to the -3/2 power .
==> (A number) to the -3/2 power means 1/(the number to the +3/2 power).
==> The 3/2 power means either find the number's square root and then cube it,
or else find the number's cube and then square root it.
w = 1 / (64 to the 3/2 power) = 1 / 512 .
That's choice ' B '.
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