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how to do I change lny = -1/2lnx + 6.8 into the form y = Ax^b ???
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stick an e on it
some use of ( ) to clean up the equation might be useful ...
ummm...... can you expand on that???? lets say I have this: lny = lnx^-1/2 + 6.8 Now what??
lny=lnx^(-0.5)+6.8 y=x^-0.5*e^6.8 Let e^6.8 be A y=Ax^-0.5
bases and logs are inverses, they cancel each other out
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lny = lnx^-1/2 + 6.8 e^(lny = lnx^-1/2 + 6.8 ) y = x^-1/2 * e^(6.8)
ohhhhhhh, zed, that makes sense.
and you too, amistre64 :D
\begin{array}l\color{red}{\text{y}}\color{orange}{\text{a}}\color{#9c9a2e}{\text{y}}\color{green}{\text{!}}\color{blue}{\text{!}}\color{purple}{\text{ }}\color{purple}{\text{:}}\color{red}{\text{)}}\color{orange}{\text{}}\end{array}
^lol
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