2^х+1 + 2^х-1 +2^х =28
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:
\[ \boxed{3} \]
