Hi Chris,
To solve for y you need to use two of the properties of logarithms. They are
log(a b) = log(a) + log(b) and
log(b^{a}) = a log(b)
The second of these properties I can use immediately to write
0.3 log_{e}x = log_{e}(x^{0.3} )
and hence
log_{e} y = 0.3log_{e}x+1.2 = log_{e}(x^{0.3} ) + 1.2
I could now use the first property of logarithms I mentioned above if the number 1.2 were a logarithm. Is there a number a so that 1.2 = log_{e}(a)? If so you can write
log_{e} y = log_{e}(x^{0.3} ) + 1.2 = log_{e}(x^{0.3} ) + log_{e}(a) = log_{e}(a x^{0.3})
and hence
y = a x^{0.3}
What is a?
Penny
