Are these base 10 logs?
Do you mean
y = log (2x)^(1/3)
If so then it
y = (1/3) log (2x)
x = (1/3) log (2y)
3 x = log 2 y
10^(3x) = 2 y
y = (1/2)(1000)(10^x)
y = 500 (10^x)
I am sorry, base 3..
log base 3 of (2x)
y = log(3)2x means that
3^y = 2x, so
x = (1/2)*3^y is the inverse function
You could also write it in inverse function notation as
f^-1(x) = (1/2)*3^x
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