In the equation 2^(x+5) = 2 + 2^x, what is the value of x?

In the equation 2^(x+5) = 2 + 2^x, what is the value of x?

• A. 4
• B. 1
• C. -1
• D. -5

Answer: (C)

To solve the equation \(2^{(x+5)} = 2 + 2^x\), we can start by rewriting the left-hand side. The term \(2^{(x+5)}\) can be expressed as \(2^5 \cdot 2^x\), which simplifies to \(32 \cdot 2^x\). This gives us the new equation: \[ 32 \cdot 2^x = 2 + 2^x \] Next, we can isolate the \(2^x\) term. We do this by subtracting \(2^x\) from both sides: \[ 32 \cdot 2^x - 2^x = 2 \] Factoring out \(2^x\) from the left-hand side results in: \[ (32 - 1) \cdot 2^x = 2 \] This simplifies to: \[ 31 \cdot 2^x = 2 \] Next, we can divide both sides by 31: \[ 2^x = \frac{2}{31} \] To solve for \(x\), we take the logarithm

To solve the equation (2^{(x+5)} = 2 + 2^x), we can start by rewriting the left-hand side. The term (2^{(x+5)}) can be expressed as (2^5 \cdot 2^x), which simplifies to (32 \cdot 2^x).

This gives us the new equation:

[

32 \cdot 2^x = 2 + 2^x

]

Next, we can isolate the (2^x) term. We do this by subtracting (2^x) from both sides:

[

32 \cdot 2^x - 2^x = 2

]

Factoring out (2^x) from the left-hand side results in:

[

(32 - 1) \cdot 2^x = 2

]

This simplifies to:

[

31 \cdot 2^x = 2

]

Next, we can divide both sides by 31:

[

2^x = \frac{2}{31}

]

To solve for (x), we take the logarithm