Answer:
Step-by-step explanation:
To solve the equation \(2^{3x - 5} \cdot a^{x - 2} = 2^{x - 2} \cdot a^{1 - x}\), we can start by simplifying the equation using exponent properties.
First, we can rewrite \(2^{x - 2}\) as \(\frac{2^x}{2^2}\), and \(a^{1 - x}\) as \(\frac{a}{a^x}\), using the rules of exponents:
\[2^{3x - 5} \cdot a^{x - 2} = \frac{2^x}{4} \cdot \frac{a}{a^x}\]
Next, we can multiply both sides of the equation by 4 to eliminate the fraction:
\[4 \cdot 2^{3x - 5} \cdot a^{x - 2} = 2^x \cdot \frac{a}{a^x}\]
Now, we can rewrite \(4\) as \(2^2\) and use the properties of exponents to combine the terms:
\[2^2 \cdot 2^{3x - 5} \cdot a^{x - 2} = 2^x \cdot \frac{a}{a^x}\]
\[2^{2 + 3x - 5} \cdot a^{x - 2} = 2^x \cdot \frac{a}{a^x}\]
\[2^{3x - 3} \cdot a^{x - 2} = 2^x \cdot \frac{a}{a^x}\]
Now that both sides have the same base (2), we can equate the exponents:
\[3x - 3 + x - 2 = x + 1\]
\[4x - 5 = x + 1\]
\[3x = 6\]
\[x = 2\]
Finally, substitute \(x = 2\) back into the original equation to find the value of \(a\):
\[2^{3(2) - 5} \cdot a^{2 - 2} = 2^{2 - 2} \cdot a^{1 - 2}\]
\[2^1 \cdot a^0 = 2^0 \cdot a^{-1}\]
\[2 \cdot 1 = 1 \cdot \frac{1}{a}\]
\[2 = \frac{1}{a}\]
\[a = \frac{1}{2}\]
So, the solution to the equation is \(x = 2\) and \(a = \frac{1}{2}\).
