To solve the equation [tex]\( 4 \times 3^{x+1} - 9^x = 27 \)[/tex], let's break down the steps one by one.
1. Express all terms with a common base:
Notice that 9 and 27 can be written as powers of 3:
[tex]\[
9 = 3^2 \quad \text{and} \quad 27 = 3^3.
\][/tex]
Also, rewrite [tex]\( 3^{x+1} \)[/tex] in an alternative form:
[tex]\[
3^{x+1} = 3^x \cdot 3.
\][/tex]
Now, substitute these into the original equation:
[tex]\[
4 \cdot (3^x \cdot 3) - (3^2)^x = 3^3.
\][/tex]
2. Simplify the equation:
Simplify the terms we substituted:
[tex]\[
4 \cdot 3^{x+1} = 4 \cdot 3 \cdot 3^x = 12 \cdot 3^x,
\][/tex]
and
[tex]\[
(3^2)^x = 3^{2x}.
\][/tex]
So, the equation now looks like:
[tex]\[
12 \cdot 3^x - 3^{2x} = 27.
\][/tex]
3. Rewrite the equation with common terms:
Observe that both sides are now written with base 3. Let [tex]\( y = 3^x \)[/tex]. Thus, the equation becomes:
[tex]\[
12y - y^2 = 27.
\][/tex]
4. Form a standard quadratic equation:
Rearrange the equation to form a standard quadratic equation:
[tex]\[
-y^2 + 12y - 27 = 0.
\][/tex]
Multiply the equation by -1 to make it easier to solve:
[tex]\[
y^2 - 12y + 27 = 0.
\][/tex]
5. Solve the quadratic equation:
Use the quadratic formula [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -12 \)[/tex], and [tex]\( c = 27 \)[/tex].
[tex]\[
y = \frac{12 \pm \sqrt{144 - 108}}{2} = \frac{12 \pm \sqrt{36}}{2} = \frac{12 \pm 6}{2}.
\][/tex]
This yields two solutions:
[tex]\[
y = \frac{12 + 6}{2} = 9 \quad \text{and} \quad y = \frac{12 - 6}{2} = 3.
\][/tex]
Recall [tex]\( y = 3^x \)[/tex], so we have:
[tex]\[
3^x = 9 \quad \text{and} \quad 3^x = 3.
\][/tex]
6. Solve for [tex]\( x \)[/tex]:
For [tex]\( 3^x = 9 \)[/tex]:
[tex]\[
3^x = 3^2 \implies x = 2.
\][/tex]
For [tex]\( 3^x = 3 \)[/tex]:
[tex]\[
3^x = 3^1 \implies x = 1.
\][/tex]
Thus, the solutions to the equation [tex]\( 4 \times 3^{x+1} - 9^x = 27 \)[/tex] are:
[tex]\[
x = 1 \quad \text{and} \quad x = 2.
\][/tex]
