Solve for k in the equation $\frac{1}{8}^{k+2}$ = 1

Analysis

The equation given is an exponential equation: $\left(\frac{1}{8}\right)^{k+2} = 1$. To solve for $k$, we need to express both sides of the equation with the same base or use the property of exponents that states any non-zero number raised to the power of zero equals 1 ($x^0 = 1$).

Concept

The fundamental rule here is the Zero Exponent Rule: $a^0 = 1$ (where $a \neq 0$).

This means that for any base (in this case, $\frac{1}{8}$), if the result is 1, the exponent must be equal to 0.

Step-by-Step Solution

1. Set up the equation: $$\left(\frac{1}{8}\right)^{k+2} = 1$$

2. Apply the Zero Exponent Rule: Since any number raised to the power of 0 is 1, we can rewrite the right side of the equation as: $$\left(\frac{1}{8}\right)^{k+2} = \left(\frac{1}{8}\right)^0$$

3. Equate the exponents: Because the bases are the same ($\frac{1}{8}$), we can equate the exponents: $$k + 2 = 0$$

4. Solve for $k$: Subtract 2 from both sides: $$k = -2$$

Conclusion

By setting the exponent to zero, we find that $k = -2$. This matches option C.

Exam Tip

Whenever you see an exponential equation equal to 1, don't waste time trying to find complex logarithms or common bases. Immediately set the entire exponent expression to zero. It is a classic JAMB "shortcut" question designed to test if you know the zero exponent rule!