The general exponential function

The curve $y=a^x$ for a positive base is already drawn with powers, roots, and logarithms. The Euler e-function is the case $a=e$, and the natural logarithm reads that case backwards. Is every other positive base a new curve, or the same e-curve read at a scaled input?

Same height at a scaled input

The base stays positive and the input stays real. One curve is $y=a^x$. The other is $y=e^t$. Drag the base and the input. The point on $y=a^x$ and the point on the e-curve at $t=x\ln a$ sit at the same height. Above 1 the power rises to the right. Below 1 it falls. At 1 it stays at height 1. In every case the power passes through $(0,1)$.

That common height is the definition. For every positive real number $a$ and every real number $x$,

$a^x=e^{x\ln a}.$

When $a=1$, $\ln a=0$, so the right side is $e^0=1$. The definition still holds, and the graph is the horizontal line at height 1.

The exponent is a quotient

Now keep $a>0$ and $a\ne 1$. Then $\ln a\ne 0$, so the scaled e-curve is continuous and strictly monotone. For every $y>0$ the equation $y=a^x$ has exactly one real solution. That solution is written

$x={\log }_{a}y.$

Swapping the letters gives the inverse $y={\log }_{a}x$. The swap is the same reflection across the diagonal that turns $y=e^x$ into the natural logarithm.

The natural logarithm is the scale in the definition, so this exponent is a quotient. Drag a positive height and a base. The picture keeps the base off 1. Pass through 1 and the curve jumps from a falling logarithm to a rising one. The point on $y={\log }_{a}x$ has height $\ln y/\ln a$.

${\log }_{a}y=\frac{\ln y}{\ln a}.$

When $a>1$, $\ln a>0$. When $0<a<1$, $\ln a<0$. That sign is why the power rose or fell in the first picture, and why this logarithm does the same. At $a=1$ every height of $a^x$ is 1, so a positive height other than 1 has no real exponent, and the quotient has denominator 0.

The remaining laws for powers of a positive base stay with powers, roots, and logarithms.

Fixed at 1, and heights multiply

Let $a>0$. A continuous function $f$ from the reals to the reals which satisfies

$f(x+y)=f(x)f(y)$

for all real $x$ and $y$, and which takes the value $f(1)=a$, is the only such function. It is $f(x)=a^x$.

The base stays positive, and it may be 1. Drag the two inputs. The product of the heights equals the height at the sum. The extra point is $(1,a)$. At base 1 every height is 1, and the product of those heights is still 1.

Fixed at the base, and heights add

Let $a>0$ with $a\ne 1$. A continuous function $g$ from the positive reals to the reals which satisfies

$g(xy)=g(x)+g(y)$

for all positive $x$ and $y$, and which takes the value $g(a)=1$, is the only such function. It is $g(x)={\log }_{a}x$.

The inputs stay positive and the base stays above 1, starting at 2 rather than at $e$. Drag the two inputs. The sum of the heights equals the height at the product. The extra point sits at the base, at height 1.

A positive base is the Euler e-function read at an input scaled by the natural logarithm. The general logarithm needs that base to be different from 1, and then the exponent is the quotient of two natural logarithms. Among continuous functions, the product rule together with the value $a$ at 1 leaves only $a^x$, and the sum rule together with the value 1 at the base leaves only ${\log }_{a}x$.