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Calculus 1 Solving logarithmic equation

  1. Jul 3, 2011 #1
    Solve for b,
    f(x)=b^x

    Here is where I am so far
    lny=lnb^x
    lny=xlnb^x
    y'/y= lnb
    e^y'/y= e^lnb
    b= e^y'/y

    My question is, can I simplify the exponent y'/y any further, or is my answer good here?
    Thank you
     
  2. jcsd
  3. Jul 3, 2011 #2
    You say to solve for b in the function [itex]f\left(x\right) = b^{x}[/itex]. From the answer provided you solve for b in terms of [itex]f\left(x\right)[/itex] and [itex]f'\left(x\right)[/itex]. Does the original problem ask for you to solve for b in terms of these functions? It is possible to solve for b in terms of only [itex]x[/itex] and [itex]f\left(x\right)[/itex].
     
  4. Jul 3, 2011 #3
    It does not require taking the derivative, that was simply the first idea I had. If there is an alternate way, please show me
     
  5. Jul 3, 2011 #4
    Look at the second line of the work that you have already provided. It should actually be [itex]ln\left(y\right) = xln\left(b\right)[/itex]. If x is not equal to zero, then you can divide both sides by x. The rest of the simplification is left to you.
     
  6. Jul 3, 2011 #5

    SammyS

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    Rather than using logs, take both sides of the equation, y=bx, to the 1/x power.
     
  7. Jul 4, 2011 #6
    thanks, I made the problem so much harder than I needed to. The redirect helped.
     
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