the solution set of f'(x)>g'(x) where f(x) = 1/2*5^2x+1 and g(x) = 5^x + 4xlog5

To determine the solution set of the inequality \( f'(x) > g'(x) \) for the functions \( f(x) = \frac{1}{2} \cdot 5^{2x} + 1 \) and \( g(x) = 5^x + 4x \log 5 \), we first need to compute the derivatives of both functions and then compare them.

Step 1: Finding the Derivative of f(x)

Let's start by differentiating \( f(x) \). The function can be broken down as follows:

• The term \( \frac{1}{2} \cdot 5^{2x} \) is an exponential function.
• The constant \( 1 \) has a derivative of \( 0 \).

Using the chain rule and the fact that the derivative of \( a^{bx} \) is \( a^{bx} \cdot b \ln(a) \), we find:

f'(x) = \frac{1}{2} \cdot 2 \cdot 5^{2x} \cdot \ln(5) = 5^{2x} \cdot \ln(5)

Step 2: Finding the Derivative of g(x)

Next, we differentiate \( g(x) \):

• The term \( 5^x \) differentiates to \( 5^x \cdot \ln(5) \).
• For \( 4x \log 5 \), we apply the product rule: the derivative is \( 4 \log 5 \).

Putting this all together, we have:

g'(x) = 5^x \cdot \ln(5) + 4 \log(5)

Step 3: Setting Up the Inequality

Now that we have both derivatives, we can set up the inequality:

5^{2x} \cdot \ln(5) > 5^x \cdot \ln(5) + 4 \log(5)

Since \( \ln(5) > 0 \), we can divide both sides by \( \ln(5) \) without changing the direction of the inequality:

5^{2x} > 5^x + \frac{4 \log(5)}{\ln(5)}

Step 4: Simplifying the Inequality

Let's define \( y = 5^x \). Then, we can rewrite the inequality:

y^2 > y + \frac{4 \log(5)}{\ln(5)}

Rearranging gives us:

y^2 - y - \frac{4 \log(5)}{\ln(5)} > 0

Step 5: Solving the Quadratic Inequality

To solve this quadratic inequality, we first find the roots of the corresponding equation:

y^2 - y - \frac{4 \log(5)}{\ln(5)} = 0

Using the quadratic formula, we find:

y = \frac{1 \pm \sqrt{1 + 4 \cdot \frac{4 \log(5)}{\ln(5)}}}{2}

This gives us two roots. Let's denote them as \( y_1 \) and \( y_2 \). The inequality \( y^2 - y - \frac{4 \log(5)}{\ln(5)} > 0 \) will be satisfied outside the interval defined by these roots:

• For \( y < y_1 \) or \( y > y_2 \)

Step 6: Converting Back to x

Recall that \( y = 5^x \). We must convert our solution back to \( x \) values:

• If \( y < y_1 \), then \( 5^x < y_1 \) implies \( x < \log_5(y_1) \).
• If \( y > y_2 \), then \( 5^x > y_2 \) implies \( x > \log_5(y_2) \).

Final Thoughts

Thus, the solution set for the inequality \( f'(x) > g'(x) \) can be expressed as:

x < \log_5(y_1) \text{ or } x > \log_5(y_2)

In summary, by finding the derivatives and solving the resulting quadratic inequality, we can determine the intervals of \( x \) where \( f'(x) \) exceeds \( g'(x) \). If you have further questions or need clarification on any steps, feel free to ask!