Answer:
Step-by-step explanation:
Let's solve the equation \(2^{3x - 5} \cdot a^{x - 2} = 2^{x - 2} \cdot a^{1 - x}\).
First, we can simplify the equation by dividing both sides by \(2^{x - 2} \cdot a^{x - 2}\) to eliminate the common factors:
\[\frac{2^{3x - 5} \cdot a^{x - 2}}{2^{x - 2} \cdot a^{x - 2}} = \frac{2^{x - 2} \cdot a^{1 - x}}{2^{x - 2} \cdot a^{x - 2}}\]
Simplifying further:
\[2^{3x - 5 - (x - 2)} = \frac{a^{1 - x}}{a^{x - 2}}\]
\[2^{2x - 3} = \frac{a^{1 - x}}{a^{x - 2}}\]
Now, we can simplify the right side of the equation using the properties of exponents:
\[2^{2x - 3} = \frac{a^{1 - x}}{a^{x - 2}} = a^{1 - x - (x - 2)} = a^{3 - 2x}\]
So, our equation becomes:
\[2^{2x - 3} = a^{3 - 2x}\]
To solve for \(x\) and \(a\), we can take the logarithm of both sides. Let's take the logarithm base 2:
\[\log_2(2^{2x - 3}) = \log_2(a^{3 - 2x})\]
\[2x - 3 = (3 - 2x) \log_2(a)\]
Next, let's isolate \(x\) by moving the terms involving \(x\) to one side of the equation:
\[2x + 2x = 3 + 3\log_2(a)\]
\[4x = 3(1 + \log_2(a))\]
Finally, solve for \(x\):
\[x = \frac{3(1 + \log_2(a))}{4}\]
Now, to find the value of \(a\), we can substitute \(x\) back into one of the original equations. Let's use the equation \(2^{2x - 3} = a^{3 - 2x}\):
\[2^{2\left(\frac{3(1 + \log_2(a))}{4}\right) - 3} = a^{3 - 2\left(\frac{3(1 + \log_2(a))}{4}\right)}\]
This equation can be solved numerically for \(a\), as it involves a logarithmic term. Without specific values for \(a\) or \(\log_2(a)\), we can't solve for \(a\) algebraically. If you have specific values for \(a\) or \(x\), we can further compute the solution.
