First, I'll answer the question posed by Benjamin Dickman: Solving problems with limitations is good practice for working with algebraic structures that do not have analogous functions. For example, solving $5^x \equiv 326 \mod 331$ is a situation where the log button on your calculator isn't going to help at all. (Neither is the bisection method.) So you have to be able to solve it using techniques that can translate to the system you're working in. It's good to force students to practice with these constraints so that (1) they appreciate the readily available techniques of $\mathbb{R}$ and (2) prepare for future topics where many algebraic systems are limited to $+$, $-$, $\times$, $\div$, and integer exponentiation.
Second, I'll answer the original question: You motivate them by showing them examples of what I just mentioned above. For example, a significant portion of cryptography depends on the fact that logs are hard/impossible in discrete systems. If you can introduce modular arithmetic, that would certainly help motivate the problem. But if you've done any matrix algebra you can see that (without diagonalization and power series), problems there need to be solved with $+$, $-$, $\times$, matrix inversion, and integer exponentiation.
Last, I'll solve the problem: (This is a method known as Shank's Algorithm. It uses the fact that continued fractions provide the quickest and best approximations to irrational values.)
We know that $5^3 < 326 < 5^4$. So let $x=3+x_1$ for some $0<x_1<1$. Now divide by $5^3$ to get $5^{x_1} = 2.608$.
Since $2.608^1 < 5 < 2.608^2$, then $x_1 = \frac{1}{1+x_2}$ for some $0<x_2<1$. Divide again $5/2.608^1 = 1.91717791...$.
Since $1.91717791 < 2.608 < 1.91717791^2$, then $x_2 = \frac{1}{1+x_3}$ for some $0<x_3<1$. Divide: $2.608/1.91717791=1.36033280...$.
Since $1.36033280^2 < 1.91717791 < 1.36033280^3$, then $x_3 = \frac{1}{2+x_4}$ for some $0<x_4<1$. Divide: $1.91717791/1.36033280^2 = 1.03602939...$.
Since $1.03602939^8 < 1.36033280 < 1.03602939^9$, then $x_4 = \frac{1}{8+x_5}$ for some $0 < x_5<1$. Divide: $1.36033280/1.03602939^8 =1.02486949...$.
And so on. At any point you can just set $x_i = 0$ and stop the process. Doing so gives the following sequence of rational numbers: $3,4,\frac{7}{2},\frac{18}{5},\frac{151}{42},\frac{169}{47},\frac{489}{136},...$. The theory of continued fractions says that the error in these approximating fractions is bounded by 1 over the square of the next denominator. Since $\frac{1}{47^2} < .0005$, then $3+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{8}}}} =\frac{151}{42} = 3.595...$ solves the problem.