Solve: $\left(\frac{1}{4}\right)^{3 z-1}=16^{z+2} \cdot 64^{z-2}$
$z =$
To solve the given exponential equation \(\left(\frac{1}{4}\right)^{3z-1} = 16^{z+2} \cdot 64^{z-2}\), we start by expressing all the bases as powers of a common base. In this case, 4 is a common base since \(\frac{1}{4} = 4^{-1}\), \(16 = 4^2\), and \(64 = 4^3\). Substituting these values into our equation, we get:
\[(4^{-1})^{3z-1} = (4^2)^{z+2} \cdot (4^3)^{z-2}\]
Now, we use the law of exponents \((a^m)^n = a^{mn}\) to simplify each term.
For the left side, we have:
\[(4^{-1})^{3z-1} = 4^{-(3z-1)} = 4^{-3z+1}\]
For the right side, we first simplify the power terms:
\[(4^2)^{z+2} = 4^{2(z+2)} = 4^{2z+4}\]
\[(4^3)^{z-2} = 4^{3(z-2)} = 4^{3z-6}\]
Combining these using the product rule \(a^m \cdot a^n = a^{m+n}\), we get:
\[4^{2z+4} \cdot 4^{3z-6} = 4^{(2z+4) + (3z-6)} = 4^{5z-2}\]
Equating the exponents on both sides of the base-4 expression \(4^{-3z+1} = 4^{5z-2}\), we obtain the following linear equation:
\[-3z + 1 = 5z - 2\]
To solve for \(z\), add \(3z\) to both sides and add 2 to both sides:
\[1 + 2 = 5z + 3z\]
\[3 = 8z\]
\[z = \frac{3}{8}\]
The solution for the given exponential equation is \(z = \frac{3}{8}\).
Exponential equations are essential in many fields, including finance and science. For instance, if you are looking at the decay of a substance or the growth of an investment, you might encounter equations where the variables are in the exponents. Solving these helps us determine how much time it takes for a substance to reduce to a certain level or how much interest a bank account will generate over time. Mastering this skill allows you to model real-world phenomena accurately.