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??

Advanced Algebra  typo? 
MathMate
(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
So what do i put as the answer? is 3 the esponential function then?

Advanced Algebra 
MathMate
"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. 
Advanced Algebra 
Kenya
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
So the second solution would be appropriate. BUT... do make sure you understand how the solution was obtained, promise?

Advanced Algebra 
Anonymous
I do! thank you so much!!

Advanced Algebra :) 
MathMate
You're welcome!
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