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

We are given the equation:

\[ \left(\frac{1}{2}\right)^{t - 1} = 64 \]

To solve for \( t \), let's first write \( 64 \) as a power of \( 2 \):

\[ 64 = 2^6 \]

But the left side of our equation involves \( \frac{1}{2} \), which is \( 2^{-1} \). So, we can also write \( 64 \) as a power of \( \frac{1}{2} \):

Recall that \( \left(\frac{1}{2}\right)^{-6} = 2^6 = 64 \), because a negative exponent means the reciprocal.

So, we have: \[ \left(\frac{1}{2}\right)^{t - 1} = \left(\frac{1}{2}\right)^{-6} \]

When the bases are the same, the exponents must be equal. Therefore:

\[ t - 1 = -6 \]

Now solve for \( t \): \[ t = -6 + 1 \] \[ t = -5 \]

Key idea: Negative exponents mean "take the reciprocal," and rewriting 64 as a power of \(\frac{1}{2}\) allows us to compare the exponents directly.

Conclusion: The value of \( t \) that satisfies the equation is \( -5 \).