In this section we will develop techniques for solving equations involving exponential functions. Consider the equation \(2^{x} = 128\). After a moment’s calculation, we find \(128 = 2^{7}\), so we have \(2^{x} = 2^{7}\). The one-to-one property of exponential functions, detailed in Theorem explogsonetoone, tells us that \(2^{x} = 2^{7}\) if and only if \(x=7\). This means that not only is \(x=7\) a solution to \(2^{x} = 2^{7}\), it is the only solution.

Now suppose we change the problem ever so slightly to \(2^{x} = 129\). We could use one of the inverse properties of exponentials and logarithms listed in Theorem invpropslogs to write \(129 = 2^{\log _{2}(129)}\). We’d then have \(2^{x} = 2^{\log _{2}(129)}\), which means our solution is \(x = \log _{2}(129)\).

After all, the definition of \(\log _{2}(129)\) is ‘the exponent we put on \(2\) to get \(129\).’ Indeed we could have obtained this solution directly by rewriting the equation \(2^{x} = 129\) in its logarithmic form \(\log _{2}(129) = x\). Either way, in order to get a reasonable decimal approximation to this number, we’d use the change of base formula, Theorem changeofbase, to give us something more calculator friendly. Typically this means we convert our answer to base 10 or base \(e\), and we choose the latter: \(\log _{2}(129) = \frac {\ln (129)}{\ln (2)} \approx 7.011\).

Still another way to obtain this answer is to ‘take the natural log’ of both sides of the equation. Since \(f(x) = \ln (x)\) is a function, as long as two quantities are equal, their natural logs are equal.

We then use the Power Rule to write the exponent \(x\) as a factor then divide both sides by the constant \(\ln (2)\) to obtain our answer.

\[ \begin{array}{rclr} 2^{x} & = & 129 & \\ \ln \left (2^{x}\right ) & = & \ln (129) & \text {Take the natural log of both sides.} \\ x \ln (2) & = & \ln (129) & \text {Power Rule} \\ x & = &\frac {\ln (129)}{\ln (2)} & \\ \end{array}\]

We summarize our two strategies for solving equations featuring exponential functions below.

Steps for Solving an Equation involving Exponential Functions
  1. Isolate the exponential function.
    1. If convenient, express both sides with a common base and equate the exponents.
    2. Otherwise, take the natural log of both sides of the equation and use the Power Rule.

Note that verifying our solutions to the equations in Example expeqnsex1 analytically holds great educational value, since it reviews many of the properties of logarithms and exponents in tandem.

For example, to verify our solution to \(2000 = 1000 \cdot 3^{-0.1 t}\), we substitute \(t = -\frac {10\ln (2)}{\ln (3)}\) and check:

\[ \begin{array}{rclr} 2000 & \stackrel {?}{=} & 1000 \cdot 3^{-0.1 \left (-\frac {10\ln (2)}{\ln (3)}\right )} & \\ 2000 & \stackrel {?}{=} & 1000 \cdot 3^{\frac {\ln (2)}{\ln (3)}} & \\ 2000 & \stackrel {?}{=} & 1000 \cdot 3^{\log _{3}(2)} & \text {Change of Base}\\ 2000 & \stackrel {?}{=} & 1000 \cdot 2 & \text {Inverse Property}\\ 2000 & \stackrel {\checkmark }{=} & 2000 & \\ \end{array}\]

We strongly encourage the reader to check the remaining equations analytically as well.

Since exponential functions are continuous on their domains, the Intermediate Value Theorem IVT applies. This allows us to solve inequalities using sign diagrams as demonstrated below.

We note here that while sign diagrams will always work for solving inequalities involving exponential functions, as we’ve seen previously, there are circumstances in which we can short-cut this method.

For example, consider number canuselogsex from Example expineq above: \(2^{x^2-3x} - 16 \geq 0\). Since the base \(2>1\), \(\log _{2}(x)\) is an increasing function meaning it preserves inequalities.

We can use this to our advantage in this case and eliminate the exponential from the inequality altogether:

\[ \begin{array}{rclr} 2^{x^2-3x} - 16 & \geq & 0 & \\ 2^{x^2-3x} & \geq & 16 & \\ \log _{2}\left (2^{x^2-3x} \right ) & \geq & \log _{2}(16) & \text {$f(x) = \log _{2}(x)$ is increasing so if $b \geq a$, $\log _{2}(b) \geq \log _{2}(a)$. } \\ x^2 - 3x & \geq & 4 &\\ \end{array} \]

Hence, we’ve reduced our given inequality to \(x^2-3x \geq 4\). As seen in Section QuadraticFunctions, we can solve this inequality by completing the square, graphing, or a sign diagram, whichever strikes the reader’s fancy.

Our next example is a follow-up to Example exptempex in Section ExponentialFunctions.

We note that, once again, we can short-cut the sign diagram in Example coffeewarmerex to solve \(70 + 90 e^{-0.1 t} > 100\). Since \(\ln (x)\) is increasing, it preserves inequality. This means we can solve this inequality as follows.

\[ \begin{array}{rclr} 70 + 90 e^{-0.1 t} & > & 100 & \\ 90 e^{-0.1 t} & > & 30 & \\ e^{-0.1 t} > & \frac {1}{3} & \\ \ln \left ( e^{-0.1 t} \right ) & > & \ln \left ( \frac {1}{3} \right ) & \text {$f(x) = \ln (x)$ is increasing so if $b \geq a$, $\ln (b) \geq \ln (a)$. } \\ -0.1 t & > & - \ln (3) & \text {$\ln \left ( \frac {1}{3} \right ) = \ln \left (3^{-1} \right ) = - \ln (3)$.} \\ t & < & \frac {-\ln (3)}{-0.1} = 10 \ln (3) & \\ \end{array} \]

Since we are given \(t \geq 0\), we arrive at the same answer \(0 \leq t < 10\ln (3)\) or \([0, 10 \ln (3))\).

Note the importance, once again, of having a base larger than \(1\) so that the corresponding logarithmic function is increasing. We can still adapt this strategy to exponential functions whose base is less than \(1\), but we need to remember the corresponding logarithmic function is decreasing so it reverses inequalities.

Our last example uses the tools of this section along with those developed in Section AppDerivatives.

Note that the graph of \(f(x) = 3xe^{-x}\) produced by desmos in Example exponentialcurvesketchingex suggests the graph of \(f\) has a horizontal asymptote \(y=0\) as \(x \rightarrow \infty \).

Indeed, it is the case that \(\lim _{x \rightarrow \infty } f(x) = 0\), however if we try to reason this analytically, we get another instance of an indeterminate form. As \(x \rightarrow \infty \), \(3x \rightarrow \infty \) but \(e^{-x} \rightarrow 0\). Hence, as \(x \rightarrow \infty \), we get the indeterminate form ‘\(\infty \cdot 0\).’ Depending on how quickly the first factor approaches ‘\(\infty \)’ and how quickly the second factor approaches ‘\(0\)’, we could end up with ‘\(\infty \)’, ‘\(0\),’ or some number in between.

We’ll explore more of this phenomenon in Exercise powerexponentialgrowthex. For now, we take it as true that exponential functions dominate polynomial functions so in the above indeterminate form, the factor \(e^{-x}\) determines the end behavior of \(f\), so \(\lim _{x \rightarrow \infty } f(x) = 0\).