Certainly! Let's solve the exponential equation [tex]\( 2^{x+1} + 2^x + 2^{x-1} = 28 \)[/tex] step by step.
First, we rewrite the terms involving exponents of [tex]\(2\)[/tex] in a simplified form. Observe the exponents:
- [tex]\(2^{x+1}\)[/tex] can be rewritten as [tex]\(2 \cdot 2^x\)[/tex].
- [tex]\(2^{x-1}\)[/tex] can be rewritten as [tex]\(\frac{1}{2} \cdot 2^x\)[/tex].
Now, substitute these into the equation:
[tex]\[
2 \cdot 2^x + 2^x + \frac{1}{2} \cdot 2^x = 28
\][/tex]
Next, we notice that all terms contain [tex]\(2^x\)[/tex]. So, let's factor [tex]\(2^x\)[/tex] out of the equation:
[tex]\[
2^x (2 + 1 + \frac{1}{2}) = 28
\][/tex]
Combine the constants inside the parentheses:
[tex]\[
2 + 1 + \frac{1}{2} = 3 + \frac{1}{2} = \frac{6}{2} + \frac{1}{2} = \frac{7}{2}
\][/tex]
So the equation becomes:
[tex]\[
2^x \cdot \frac{7}{2} = 28
\][/tex]
To isolate [tex]\(2^x\)[/tex], multiply both sides of the equation by [tex]\(\frac{2}{7}\)[/tex]:
[tex]\[
2^x = 28 \cdot \frac{2}{7} = 4 \cdot 2 = 8
\][/tex]
We now have:
[tex]\[
2^x = 8
\][/tex]
Recognize that [tex]\(8\)[/tex] can be written as [tex]\(2^3\)[/tex]:
[tex]\[
2^x = 2^3
\][/tex]
Since the bases are identical, we can equate the exponents:
[tex]\[
x = 3
\][/tex]
Therefore, the solution to the equation [tex]\(2^{x+1} + 2^x + 2^{x-1} = 28\)[/tex] is:
[tex]\[
x = 3
\][/tex]
