To solve the equation ( 2^{x+1} + 2^{x-1} + 2^x = 28 ), we can first rewrite the terms with a common base. Notice that:
[
2^{x+1} = 2 \cdot 2^x
]
[
2^{x-1} = \frac{2^x}{2}
]
Now, substituting these into the equation gives:
[
2 \cdot 2^x + \frac{2^x}{2} + 2^x = 28
]
Let's denote ( 2^x ) as ( y ). Then the equation becomes:
[
2y + \frac{y}{2} + y = 28
]
Now, simplify the left side:
- Combine like terms: ( 2y + y = 3y ).
- Adding ( \frac{y}{2} ) gives:
[
3y + \frac{y}{2} = 28
]
To combine ( 3y ) and ( \frac{y}{2} ), convert ( 3y ) to have a common denominator of 2:
[
3y = \frac{6y}{2} \quad \text{(since ( 3y = \frac{6y}{2} ))}
]
So the equation becomes:
[
\frac{6y}{2} + \frac{y}{2} = 28
]
Combine the terms:
[
\frac{6y + y}{2} = \frac{7y}{2} = 28
]
Now, multiply both sides by 2 to eliminate the fraction:
[
7y = 56
]
Now, divide by 7:
[
y = 8
]
Recall that ( y = 2^x ). Therefore, we have:
[
2^x = 8
]
Since ( 8 ) can be written as ( 2^3 ), we have:
[
2^x = 2^3
]
This implies:
[
x = 3
]
Thus, the solution to the equation is:
[
\boxed{3}
]
