Exponentials and Powers

The exponential function $\exp$ is, as might be expected, closely related to the operation of taking powers. In fact, any of our functions satisfying the exponential property will have a similar relation.

To this end, consider such a function $f$ and define a positive number $b=f(1)$ . Then we can calculate $f(2)=f(1)f(1)=b^2$ , $f(3)=b^3$ , and so on. Since $f(\frac{1}{2})f(\frac{1}{2})=f(1)=b$ we see that $f(\frac{1}{2})=b^{\frac{1}{2}}$ , and similarly for all other rational numbers.

So we have a function $f(x)$ defined on all real numbers, and we have a function $b^x$ defined on all rational numbers, and where both functions are defined they agree. Since the rationals are dense in the reals (the latter being the uniform completion of the former) there can be only one continuous extension of $b^x$ to the whole real line. We’ll discard the function $f(x)$ and just write $b^x$ for this extension from now on. In particular, the function $\exp$ gives us a special number $e=\exp(1)$ , and we write $e^x=\exp(x)$ .

Like we saw before, we can use the exponential function $e^x$ to give all the other exponentials $b^x$ . We know that $b^x=e^{C_bx}$ for some constant $C_b$ , but which? If we take the natural logarithm of both sides we see that $\ln(b^x)=C_b x$ . In particular, setting $x=1$ we find $C_b=\ln(b)$ . That is, given any positive real number $b$ we can define the exponential $b^x$ as $e^{\ln(b)x}$

1 Comment »

Hey John, you’ve got a parse error in the last paragraph.

Comment by Jon | April 16, 2008 | Reply

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The Unapologetic Mathematician

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