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my physics teacher told us that for now well be solving the equation of a line such as [tex]y=kx^n[/tex] by trial and error for finding "n" using the multiplicative change in 2 points i.e [tex]x_1,y_1[/tex],[tex]x_2,y_2[/tex].
what we would do is for example 2 points (13,1) (6, 1.4):
[tex]n=\frac{y_2}{y_1}=(\frac{x_2}{x_1})^n[/tex]
[tex]n=\frac{6}{13}=(\frac{1.4}{1})^n[/tex]
so:
[tex]~0.46=1.4^?[/tex]
we'd have to use trial and error...wat id like to know is how to use logarithm to solve this, he said he might teach it but i was wondering if u guys could fill me in on how to solve this using log:
[tex]~0.46=1.4^n[/tex],
the answer for n is about -2.3...
thanks in advance
what we would do is for example 2 points (13,1) (6, 1.4):
[tex]n=\frac{y_2}{y_1}=(\frac{x_2}{x_1})^n[/tex]
[tex]n=\frac{6}{13}=(\frac{1.4}{1})^n[/tex]
so:
[tex]~0.46=1.4^?[/tex]
we'd have to use trial and error...wat id like to know is how to use logarithm to solve this, he said he might teach it but i was wondering if u guys could fill me in on how to solve this using log:
[tex]~0.46=1.4^n[/tex],
the answer for n is about -2.3...
thanks in advance