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5^(-T/2)=.20 I know the answer, but HOW does one arrive there? What rules are in play here?
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@Gralwyn
helps to notice that 0.2=1/5, so you can rewrite 5^(-T/2)=1/5 now take the log base 5 of both sides
how does one do that correctly?
do you know about logarithms?
\[5^{\frac{ -T }{ 2 }}=0.20=\frac{ 20 }{100 }=\frac{ 1 }{5 }=5^{-1}\] \[\frac{ -T }{2 }=-1,T=?\]
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Yes, but I was under the assumption that I could take the log of just one side so \[(-T/2)\log5{.20}\]
5 is supposed to be the base there.
you cannot "just take the log of one side"\[x=y\neq\log x\]
ok, so log5(-T/2)=log5?
T=-1*-2=2
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