Let's solve the equation [tex]\(2^{x-3} \cdot 2 \cdot a^{1-x} = 2^{3x-5} \cdot a^{x-2}\)[/tex].
### Step-by-Step Solution:
1. Combine the Exponents of 2:
First, notice that we can combine the terms involving the base 2 on each side.
The left side [tex]\(2^{x-3} \cdot 2\)[/tex] can be rewritten as:
[tex]\[
2^{x-3} \cdot 2^1 = 2^{(x-3) + 1} = 2^{x-2}
\][/tex]
So our rewritten equation now becomes:
[tex]\[
2^{x-2} \cdot a^{1-x} = 2^{3x-5} \cdot a^{x-2}
\][/tex]
2. Compare the Exponents of 2:
Since we have [tex]\(2^{something}\)[/tex] on both sides, we can compare the exponents:
[tex]\[
x-2 = 3x-5
\][/tex]
3. Solve for x:
To solve for [tex]\(x\)[/tex], we isolate [tex]\(x\)[/tex] by first subtracting [tex]\(x\)[/tex] from both sides:
[tex]\[
x - 2 - x = 3x - 5 - x
\][/tex]
Simplify to:
[tex]\[
-2 = 2x - 5
\][/tex]
Next, add 5 to both sides:
[tex]\[
-2 + 5 = 2x
\][/tex]
Simplifies to:
[tex]\[
3 = 2x
\][/tex]
Finally, divide by 2:
[tex]\[
x = \frac{3}{2}
\][/tex]
So the solution to the equation [tex]\(2^{x-3} \cdot 2 \cdot a^{1-x} = 2^{3x-5} \cdot a^{x-2}\)[/tex] is:
[tex]\[
x = \frac{3}{2}
\][/tex]
