y= k(x+1) to the power of n

find approximate values for k and n given that

x 4 8 15

y 4.45 4.60 4.8

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- May 5th 2009, 11:14 PMscouserreduction of a relationship to a linear law
y= k(x+1) to the power of n

find approximate values for k and n given that

x 4 8 15

y 4.45 4.60 4.8 - May 6th 2009, 01:57 AMGrandadLogs
Hello scouserTake logs of both sides:

$\displaystyle \log y = \log\Big(k(x+1)^n\Big)$

$\displaystyle = \log k +n\log(x+1)$

Plot the graph of $\displaystyle \log y$ against $\displaystyle \log(x+1)$, using the three pairs of values you're given. Draw the best straight line and read off the gradient and intercept.

Gradient = $\displaystyle n \approx 0.065$

Intercept = $\displaystyle \log k$, which gives $\displaystyle k \approx 4$

Grandad