Convert y=ae^lnx to y=ab^x

\(\displaystyle e^{(0.1539 \cdot x)} \ \ = \ \ \left [{e^{0.1539}\right ]}^x}\)

Now continue.....
 
Evaluate couple of "y"s for different values of "x" - using both formulae - and see if those are correct.
 
Using a x value of 20 I came up with 387.38 and 415.602. They seem far of from the rest of my data.
 
I get:

17.8013*1.1665^20 = 387.7082716

and

17.8013*e^(0.1539*20) = 386.8731765
 
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