Consider the function g defined by g (x) = 2x - 2x, x ER.
Point A is the point on the graph of g for which the x-coordinate is In √3.
Find the equation of the tangent to the graph of g at point A.

Let's solve the problem step by step.

### Step 1: Simplify the Function \( g(x) \)
The given function is \( g(x) = 2x - 2^x \).

### Step 2: Find the Derivative of \( g(x) \)
To find the equation of the tangent line, we need the derivative of \( g(x) \) with respect to \( x \). The derivative, \( g'(x) \), gives us the slope of the tangent line at any point \( x \).

\[
g'(x) = \frac{d}{dx} (2x - 2^x)
\]

The derivative of \( 2x \) with respect to \( x \) is \( 2 \), and the derivative of \( 2^x \) with respect to \( x \) is \( 2^x \ln(2) \).

So,

\[
g'(x) = 2 - 2^x \ln(2)
\]

### Step 3: Find the Slope at Point A
Point A has an \( x \)-coordinate of \( x = \ln(\sqrt{3}) \).

To find the slope of the tangent at point A, substitute \( x = \ln(\sqrt{3}) \) into \( g'(x) \):

\[
g'(\ln(\sqrt{3})) = 2 - 2^{\ln(\sqrt{3})} \ln(2)
\]

Now, \( 2^{\ln(\sqrt{3})} \) can be simplified using the property \( a^{\ln(b)} = b^{\ln(a)} \):

\[
2^{\ln(\sqrt{3})} = (\sqrt{3})^{\ln(2)}
\]

So the slope \( m \) at point A is:

\[
m = 2 - (\sqrt{3})^{\ln(2)} \cdot \ln(2)
\]

### Step 4: Find the y-Coordinate of Point A
The \( y \)-coordinate of point A is found by substituting \( x = \ln(\sqrt{3}) \) into the original function \( g(x) \):

\[
g(\ln(\sqrt{3})) = 2\ln(\sqrt{3}) - (\sqrt{3})^{\ln(2)}
\]

Thus, point A is \( \left(\ln(\sqrt{3}), 2\ln(\sqrt{3}) - (\sqrt{3})^{\ln(2)}\right) \).

### Step 5: Write the Equation of the Tangent Line
The equation of the tangent line at point A can be written as:

\[
y - y_1 = m(x - x_1)
\]

Substituting the coordinates of point A and the slope:

\[
y - \left(2\ln(\sqrt{3}) - (\sqrt{3})^{\ln(2)}\right) = \left(2 - (\sqrt{3})^{\ln(2)} \cdot \ln(2)\right)\left(x - \ln(\sqrt{3})\right)
\]

### Final Answer:
The equation of the tangent to the graph of \( g(x) = 2x - 2^x \) at point \( A \) is:

\[
y = \left(2 - (\sqrt{3})^{\ln(2)} \cdot \ln(2)\right) \cdot (x - \ln(\sqrt{3})) + \left(2\ln(\sqrt{3}) - (\sqrt{3})^{\ln(2)}\right)
\]

Would you like more details on any part of the solution, or do you have any further questions?

Here are some related questions to consider:
1. How do you find the derivative of exponential functions?
2. What is the significance of the slope of a tangent line?
3. How can you simplify expressions like \( 2^{\ln(\sqrt{3})} \)?
4. How do you use the point-slope form to write the equation of a line?
5. What are the properties of logarithms that are useful in calculus?

**Tip:** When working with exponential and logarithmic functions, remember that simplifying expressions using properties of exponents and logarithms can make differentiation easier.

Learn how to find the equation of the tangent line to the graph of g(x) = 2x - 2^x at the point where x = ln(sqrt(3)). This problem involves differentiation of exponential and logarithmic functions and using the point-slope form of a line. Suitable for advanced high school students.

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