What is an exponential function in the form y=ab^x that passes through the points (1,12) and (1,36)

Question

I really need help ASAP

Answer 1

There is no exponential that goes though both (1, 12) and (1, 36). You would need a vertical line, i.e. b = Infinity.

But, in general, if you have an exponential that goes through (x0, y0) and (x1, y1), then:

y0 = f(x0) = a * b^x0

y1 = f(x1) = a * b^x1

then we can find b using:

y0 / y1 = f(x0) / f(x1) = (a * b^x0) / (a * b^x1) = b^x0 / b^x1 = b^(x0 - x1).

That is,

y0 / y1 = b^(x0 - x1)

Taking the log of both sides (say the log base e or the natural log), we find:

ln(y0 / y1) = (x0 - x1) * ln(b)

Solving for ln(b):

ln(b) = ln(y0 / y1) / (x0 - x1)

and exponentiating:

b = e^(ln(y0 / y1) / (x0 - x1))

or

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|| b = (y0/y1)^[ 1 / (x0 - x1)] ||

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Then, to find a, use:

y0 = a * b^x0 = a * (y0/y1)^[ x0 / (x0 - x1)]

And thus,

a = y0 / { (y0/y1)^[ x0 / (x0 - x1)] }

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|| a = y0 * (y0/y1)^[ -x0 / (x0 - x1)]. ||

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