To solve the equation [tex]\(2^{x+2} + \frac{2^{x+3}}{2} = 1\)[/tex], let's go through the steps in detail:
1. Simplify the equation:
Let's start with simplifying the given equation:
[tex]\[
2^{x+2} + \frac{2^{x+3}}{2} = 1
\][/tex]
Notice that [tex]\(\frac{2^{x+3}}{2}\)[/tex] can be simplified. Since [tex]\(\frac{2^{a}}{2^b} = 2^{a-b}\)[/tex], we get:
[tex]\[
\frac{2^{x+3}}{2} = 2^{(x+3) - 1} = 2^{x+2}
\][/tex]
2. Combine like terms:
Substitute the simplified term back into the equation:
[tex]\[
2^{x+2} + 2^{x+2} = 1
\][/tex]
Combine the terms on the left side:
[tex]\[
2 \cdot 2^{x+2} = 1
\][/tex]
We know that [tex]\(2 \cdot 2^{x+2}\)[/tex] can be simplified as:
[tex]\[
2^{1} \cdot 2^{x+2} = 2^{x+2+1} = 2^{x+3}
\][/tex]
So, we now have:
[tex]\[
2^{x+3} = 1
\][/tex]
3. Solve for the exponent:
Since [tex]\(1\)[/tex] can be written as [tex]\(2^0\)[/tex] (because [tex]\(2^0 = 1\)[/tex]):
[tex]\[
2^{x+3} = 2^0
\][/tex]
For the equality to hold, the exponents must be equal:
[tex]\[
x + 3 = 0
\][/tex]
4. Solve for [tex]\(x\)[/tex]:
Isolate [tex]\(x\)[/tex] by subtracting 3 from both sides:
[tex]\[
x = -3
\][/tex]
So, the solution to the equation [tex]\(2^{x+2} + \frac{2^{x+3}}{2} = 1\)[/tex] is:
[tex]\[
x = -3
\][/tex]
Hence, the value of [tex]\(x\)[/tex] is [tex]\(-3\)[/tex].
