Find the value of t for which ($\frac{1}{2}$)$^{t - 1}$ = 64

Analysis

To solve this equation, we need to express both sides of the equation with the same base. Once the bases are identical, we can equate the exponents to solve for $t$.

Concept

This is an exponential equation. The key principle is: If $a^x = a^y$, then $x = y$.

We know that:

• $\frac{1}{2}$ can be written as $2^{-1}$.
• $64$ is a power of $2$ ($2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$, so $64 = 2^6$).

Step-by-Step Solution

Step 1: Rewrite the equation using base 2. The original equation is: $(\frac{1}{2})^{t - 1} = 64$

Substitute $\frac{1}{2}$ with $2^{-1}$: $(2^{-1})^{t - 1} = 2^6$

Step 2: Apply the law of indices $(a^m)^n = a^{m \times n}$. Multiply the exponents on the left side: $2^{-1(t - 1)} = 2^6$ $2^{-t + 1} = 2^6$

Step 3: Equate the exponents. Since the bases are now the same (both are 2), we can set the exponents equal to each other: $-t + 1 = 6$

Step 4: Solve for $t$. Subtract 1 from both sides: $-t = 6 - 1$ $-t = 5$

Multiply both sides by -1: $t = -5$

Conclusion

The value of $t$ that satisfies the equation is $-5$.

Exam Tip: Whenever you see a fraction like $\frac{1}{2}, \frac{1}{3},$ or $\frac{1}{4}$ in an exponential equation, remember that they are just negative powers of the denominator (e.g., $\frac{1}{2} = 2^{-1}$, $\frac{1}{4} = 4^{-1}$ or $2^{-2}$). Converting to negative powers early makes the calculation much easier!