Let's solve each equation using logarithmic notation.

Problem (a): $5^x = 56$

To express $x$ in logarithmic form, take the logarithm of both sides with base 5:

$x = \log_5(56)$

Problem (b): $7^x = \frac{1}{2401}$

First, rewrite $\frac{1}{2401}$ as $7^{-4}$ since $2401 = 7^4$. Then we have:

$7^x = 7^{-4}$

This implies $x = -4$, or equivalently:

$x = \log_7\left(\frac{1}{2401}\right) = -4$

Problem (c): $5^x = -80$

There is no real solution for $x$ in this case, because $5^x$ (an exponential function) cannot produce a negative result. Therefore, the answer is:

$x = \text{DNE}$

Do you want more details on any of these steps, or do you have any questions?

Here are some related questions to consider:

1. How do we convert exponential equations to logarithmic form in general?
2. Why can't exponential functions with real exponents yield negative results?
3. What alternative solutions exist if complex numbers are considered for equation (c)?
4. How does the base of the logarithm affect the solution of an equation?
5. How would you solve similar equations if the base were a variable rather than a constant?

Tip: Always check if the exponential expression can logically reach the target value before solving, as some exponential equations have no solution in the real number system.