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anonymous

  • one year ago

I'm missing this conceptual connection... ln (b) is the slope at x=0 of the tangent to f(x)=b^x. What is the connection to e to that power being equal to base b?

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  1. phi
    • one year ago
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    here is an example with y= 5^x and the tangent line at (0,1) y= ln(5)x +1

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  2. phi
    • one year ago
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    I assume you know that \[ e^{\ln(b)} = b\] so we can write \[ b^x = \left( e^{\ln(b)} \right)^x =e^{x\ln(b)} \] the ln(b) is a constant. The derivative with respect to x is \[ \frac{d}{dx} b^x= \frac{d}{dx} e^{x\ln(b)} = e^{x\ln(b)} \frac{d}{dx} x\ln(b) \\ = e^{x\ln(b)}\ln(b) \\= \ln(b)\ b^x \] at x=0 we get dy/dx= ln(b) thus the tangent to the curve y= b^x at x=0 is y-1= ln(b)(x-0) or y= ln(b) x + 1

  3. anonymous
    • one year ago
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    Thank you... I do understand the derivation. I'm trying to conceptualize the significance of the rate of change at x=0 of b^x being the natural log.

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