Y = mx + c...?

How to express this in Y=Mx + c


y =ab^x


y=Ax^k


px+qv = xy





I know you have to put log.. into it, but can someone please teach me the way to do so?





Thanks!

Y = mx + c...?
Perhaps this is what you mean?


log y = log (ab^x) = log a + xlog b


If you plot log y verses x, it will be a straight line. IE define log y = Y


For the second


log y = log ax^k = log a + k log x


log y and log x are in a linear relation.


I am not sure if that is what you are looking for.
Reply:Do you mean (ab)^x? If not, log y=log ab^x


log y=log a + log b^x


log y=x log b + log a


The same way for the second question.





px+qv=xy


Divide by x the whole equation:


y=qv/x + p


log y=log (qv/x) + log p


log y=log qv - log x + log p


log y=(-1)logx + log pqv
Reply:Actually I never put it in log form because it's pretty much the same thing, but if you do want me to put it in that form thats fine anyway.


y=ab^x


y/a=b^x


log base b of y/a =x


The next one is pretty much the same thing.


log base x of y/a=k


3rd: This one has no log sign into it.
Reply:y = ab^x


put log on both sides


lg y = lg ab^x


lg y = lg a + lg b^x


lg y = x lg b + lg a


lg y = lg b (x) + lg a





y = Ax^k


put lg on both sides


lg y = lg Ax^k


lg y = lg A + lg x^k


lg y = k lg x + lg A





px + qv = xy


xy = - px - qv
Reply:I don't believe the first two are possible. Y=mx+c describes a linear function, and once you drop an exponent into the mix - ZOT! linearity is gone.

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