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Quick Question differentiating logs

  1. Oct 23, 2011 #1
    1. my function is: f(x)=log 1.5 (-.76x+305). f(x)= log base 1.5 of -.76x+305



    3. How do i differentiate it? here is what i have so far: (1/((-.76x+305)ln1.5))*(-.76/dx)
     
  2. jcsd
  3. Oct 23, 2011 #2

    Dick

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    That's pretty close. (log base 1.5)(-.76x+305) is equal to log(-.76x+305)/log(1.5), right? I assume that's what you are starting from. Now just differentiate it using the chain rule. My only question is what is the '/dx' doing in there?
     
  4. Oct 24, 2011 #3
    yes, but there seems to be truly no need for the chain rule, as log(1.5) is a number, equal to 0.176091

    so it seems i just multiply the derivative of [log base 10 (-0.76x+305)] by that number!
     
  5. Oct 24, 2011 #4
    ((-0.76)(log(e)))/((-0.76x+305)(log(1.5))

    thats the answer i think.... in case anyone is curious
     
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