Question 987414
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You might have a better time understanding this if you consider that points on the graph of some function of *[tex \Large x] are of the form *[tex \Large \left(x,\,f(x)\right)].  So if *[tex \Large x_1] is the *[tex \Large x]-coordinate of a point on the graph, then *[tex \Large f(x_1)] is the *[tex \Large y]-coordinate of that point.


From this idea we get that if *[tex \Large (1,6)] is a point on a graph of *[tex \Large f(x)\ =\ a*b^x],  Then it must be true that *[tex \Large a*b\ =\ 6].  And if *[tex \Large \left(-1,\frac{3}{2}\right)] is on the graph, then *[tex \Large \frac{a}{b}\ =\ \frac{3}{2}]


So, since *[tex \Large a\ =\ a], we can write:


*[tex \Large \frac{6}{b}\ =\ \frac{3b}{2}]


Solve for *[tex \Large b] remembering the restriction that *[tex \Large b\ >\ 0].  Then substitute into either equation and solve for *[tex \Large a]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \