APPROACH #1: Number sense

First recognize that $79^{43}$ = $79^{31} \times 79^{12}$

While $79^{31}$ might be a little bit smaller than 80^{31}, once we multiply $79^{31}$ by $79^{12}$ (a very large number), the resulting $79^{43}$ will definitely be greater than $80^{31}$

Answer: A


APPROACH #2: Exponent laws and inequality laws
Let's first start with some matching operations (see video below on this technique)

Given:
Quantity A: $79^{43}$
Quantity B: $80^{31}$

Divide both quantities by $79^{31}$ to get:

Quantity A: $\frac{79^{43}}{79^{31}}$

Quantity B: $\frac{80^{31}}{79^{31}}$


Simplify to get:
Quantity A: $79^{12}$

Quantity B: $(\frac{80}{79})^{31}$

At this point we're comparing $79^{12}$ with $(\frac{80}{79})^{31}$

First notice that $64^{12} < 79^{12}$

Now rewrite $64$ and $2^6$ to get: $(2^6)^{12} < 79^{12}$

Simplify $(2^6)^{12}$ to get: $2^{72} < 79^{12}$

Since $\frac{80}{79} < 2$, we know that $(\frac{80}{79})^{72} < 2^{72}$

Finally, since $(\frac{80}{79})^{31} < (\frac{80}{79})^{72}$, we can write: $(\frac{80}{79})^{31} < (\frac{80}{79})^{72} < 2^{72} < 79^{12}$

This tells us that: $(\frac{80}{79})^{31} < 79^{12}$

Answer: A


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