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Posted by 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? - , 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 - , Sunday, May 1, 2011 at 10:37am

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

  • Advanced Algebra - , 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 - , 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 - , 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 - , Sunday, May 1, 2011 at 2:12pm

    I do! thank you so much!!

  • Advanced Algebra :) - , Sunday, May 1, 2011 at 2:54pm

    You're welcome!

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