Question

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?

Answer 1

here is an example with y= 5^x and the tangent line at (0,1) y= ln(5)x +1

Answer 2

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

Answer 3

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.