To solve the equation [tex]\( 4 \cdot 3^{1-x} + 3^{2-x} = 63 \)[/tex], follow these steps:
1. Rewrite the equation in terms of a single exponent base, [tex]\(3\)[/tex]:
[tex]\[
4 \cdot 3^{1-x} + 3^{2-x} = 63
\][/tex]
2. Substitute [tex]\( y = 3^{-x} \)[/tex]:
[tex]\[
4 \cdot 3 \cdot y + 9 \cdot y = 63
\][/tex]
Simplify this:
[tex]\[
12y + 9y = 63
\][/tex]
Combine like terms:
[tex]\[
21y = 63
\][/tex]
3. Solve for [tex]\( y \)[/tex]:
[tex]\[
y = \frac{63}{21} = 3
\][/tex]
4. Recall that [tex]\( y = 3^{-x} \)[/tex], so:
[tex]\[
3^{-x} = 3
\][/tex]
5. Solve for [tex]\( x \)[/tex] using properties of exponents:
[tex]\[
-x = 1
\][/tex]
Therefore:
[tex]\[
x = -1
\][/tex]
So, the solution to the equation [tex]\( 4 \cdot 3^{1-x} + 3^{2-x} = 63 \)[/tex] is [tex]\( x = -1 \)[/tex].
Let's check this solution against the given multiple choice options:
a. [tex]\( x = -3 \)[/tex]
b. [tex]\( x = -1 \)[/tex]
c. [tex]\( x = 1 \)[/tex]
d. [tex]\( x = 3 \)[/tex]
The correct answer is:
[tex]\[ \boxed{x = -1} \][/tex]
