Advanced Algebra
posted by Anonymous .
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??

(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 
So what do i put as the answer? is 3 the esponential function then?

"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 doublecheck what you've copied the question correctly before you start. 
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 
So the second solution would be appropriate. BUT... do make sure you understand how the solution was obtained, promise?

I do! thank you so much!!

You're welcome!
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