How would one solve this exponent problem without a calculator?

Question

This is a problem on a non-calculator portion of an algebra 2 honors worksheet.

$$2^{24}=(3x-1)^8$$

Answer 1

Take the eighth root.

$2^3 = 3x - 1$

$9 = 3x$

$x = 3$

Another solution is -7/3 because after the final square root, we take one of the negative solutions.

There are many other solutions, of course, according to the fundamental theorem of algebra. But, they end up with x being complex numbers.

Answer 2

To find just real roots, start by taking the square root of both sides:

$$(3x-1)^4=2^{12} \text{ or } (3x-1)^4=-2^{12}$$

Since the right equation has no real solutions, we can ignore it. Next, take the square root of the left equation:

$$(3x-1)^2=2^6 \text{ or } (3x-1)^2=-2^6$$

Again, the equation on the right has no real solutions, so it can be ignored. Again take the square root of the left equation:

$$3x-1=2^3 \text{ or } 3x-1=-2^3$$

This time the equation on the right does have real solutions, so we can work with both to find the real roots.

$$ \begin{align*} 3x-1&=8\\ 3x&=9\\ x&=3\\ \\ 3x-1&=-8\\ 3x&=-7\\ x&=-\frac{7}{3} \end{align*} $$