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(ba) = a log(b)
The second of these properties I can use immediately to write
0.3 logex = loge(x0.3 )
and hence
loge y = 0.3logex+1.2 = loge(x0.3 ) + 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 = loge(a)? If so you can write
loge y = loge(x0.3 ) + 1.2 = loge(x0.3 ) + loge(a) = loge(a x0.3)
and hence
y = a x0.3
What is a?
Penny
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