Posted by **Anonymous** on Saturday, April 30, 2011 at 11:54pm.

Find the exponential function of the form y=ax^b whose graph passes through the points (-3, 1/27) and (0,1).

I don't know where to begin with this problem??

- Advanced Algebra - typo? -
**MathMate**, Sunday, May 1, 2011 at 7:10am
(0,1) means when x=0, y=1.

When x=0,

y = a*0^b = 0 except when b=0, when y is undefined.

So y cannot equal 1 when x=0.

Please check if there is a typo, for example: y=ab^x instead.

If this is the case, then

at (0,1),

1=ab^0=a*1=a

so y=b^x

at (-3,1/27)

1/27 = b^(-3)

27 = b^3;

3^3=b^3

b=3

- Advanced Algebra -
**Anonymous**, Sunday, May 1, 2011 at 10:37am
So what do i put as the answer? is 3 the esponential function then?

- Advanced Algebra -
**MathMate**, Sunday, May 1, 2011 at 11:35am
"Please check if there is a typo,..."

When you see what the question actually asks, then you can decide which way to go.

Especially in exams, you need to double-check what you've copied the question correctly before you start.

- Advanced Algebra -
**Kenya**, Sunday, May 1, 2011 at 12:25pm
Im sorry yes there was a typo, you correctly fixed it. the b should be an x so it should look like this.

y=ab^x

- Advanced Algebra -
**MathMate**, Sunday, May 1, 2011 at 1:39pm
So the second solution would be appropriate. BUT... do make sure you understand how the solution was obtained, promise?

- Advanced Algebra -
**Anonymous**, Sunday, May 1, 2011 at 2:12pm
I do! thank you so much!!

- Advanced Algebra :) -
**MathMate**, Sunday, May 1, 2011 at 2:54pm
You're welcome!

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