# Exponential Kiss

Alignments to Content Standards: F-LE.A.4 F-BF.B.5 F-IF.C.7.e

Below is a picture of the functions $f(x) = \log_b{x}$ and $g(x) = b^x$. In the application below, the base b varies between 1 and 2 (by hundredths) and pressing the "play" button will run through all possible values of b.

1. For which values of $b$ do the two graphs appear not to meet?
2. For which values of $b$ do the two graphs appear to meet in two points?
3. Check by evaluating the functions $f$ and $g$ that the two graphs meet at $x=e$ when $b = e^{\frac{1}{e}}$.

## IM Commentary

The purpose of this task is twofold: first using technology to study the behavior of some exponential and logarithmic graphs and secondly to manipulate some explicit logarithmic and exponential expressions. Although not asked in the task body, the teacher may wish to prompt students to explain why the two graphs behave as they do as the base $b$ varies: that is, a larger value of $b$ between 1 and 2 makes the exponential graph grow faster and the logarithmic graph grow more slowly as $x$ increases. These two phenomena are related to the additional fact that the functions $f(x) = e^x$ and $g(x) = \ln{x}$ are inverses of one another.

In the cases where the graphs meet in two points, it is usually challenging to calculate those points. We consider here the case where $b = \sqrt{2}$. We can check by direct calculation that $x = 2$ is a solution to $$(\sqrt{2})^x = \log_{\sqrt{2}}{x}.$$ In this particular case, the second solution to the equation can also be calculated directly and it is $x = 4$. More generally, if $b = a^{\frac{1}{a}}$, then the two graphs meet when $x = a$ (we have just checked this for the case $a = 2$). Finding the second point of intersection, however, is a challenging question!

## Solution

1. If we zoom in on the graph, we can see that the two graphs appear not to meet when $1.45 \leq b \leq 2$
2. Similarly, zooming in on the graphs reveals that they meet in two points when $1 \leq b \leq 1.44$. When $b$ takes the values between $1.44$ and $1.45$ we would need to refine the increment of $b$ in the application in order to check whether or not they meet in no points, one point, or two points.
3. If $b = e^{\frac{1}{e}}$ then $$f(x) = \log_b{x} = \log_{e^{\frac{1}{e}}}(x).$$ When $x = e$ we have

$$f(e) = \log_{e^{1/e}}(e),$$

so to evaluate $f(e)$ we find the value $a$ such that $(e^{1/e})^a=e$.  Since $\left(e^{\frac{1}{e}}\right)^e = e^{\frac{1}{e} \cdot e} = e^1 = e$ we have $f(e) = e$.

Now we compute $g(e)$: $$g(e)=\left(e^{1/e}\right)^e=e^{e\cdot 1/e}=e^1=e.$$

Checking on a calculator $e^{\frac{1}{e}}$ is between 1.44 and 1.45 which agrees with what we have found in parts (a) and (b).