If ( log _{7}^{2}, 7^{log }left(e^{x}-5right), 7^{log _{7}left(2^{2}-frac{7}{2}right)} ) are in geometric progression then the sum of the possible values

1. Break Down the Problem

We need to analyze the given three terms:

1. $\log_{7}(2)$
2. $7^{\log(e^{x}-5)}$
3. $7^{\log_{7}(2^{2}-\frac{7}{2})}$

We will determine the condition under which these terms are in geometric progression (GP). For three terms $a, b, c$, they are in GP if $b^2 = ac$.

2. Relevant Concepts

To prove the terms are in GP:

• Set $a = \log_{7}(2)$
• Set $b = 7^{\log(e^{x}-5)}$
• Set $c = 7^{\log_{7}(2^{2}-\frac{7}{2})} = 2^{2}-\frac{7}{2}$ (because $7^{\log_{7}(y)} = y$)

3. Analysis and Detail

First, we express the condition for GP: $b^2 = ac \implies \left(7^{\log(e^{x}-5)}\right)^2 = \log_{7}(2) \cdot \left(2^{2}-\frac{7}{2}\right)$ This gives us: $7^{2\log(e^{x}-5)} = \log_{7}(2) \cdot \left(2-\frac{7}{2}\right)$ Now we will further analyze $b$:

Knowing that $7^{\log(a)} = a^{\log(7)}$, we can rewrite $7^{\log(e^{x}-5)} = (e^{x}-5)^{\log(7)}$. Thus, we have: $(e^{x}-5)^{\log(7)} = \sqrt{\log_{7}(2) \cdot (2^{2}-\frac{7}{2})}$ Next, we evaluate the term: $2^{2} - \frac{7}{2} = 4 - 3.5 = 0.5$ Now, substituting this into the equation: $(e^{x}-5)^{\log(7)} = \sqrt{\log_{7}(2) \cdot 0.5}$

4. Verify and Summarize

Now we need to resolve: $b^2 = ac \quad \text{and evaluate: } \log_{7}(2) \text{ is a constant value. }$ Once we perform this, we can derive $e^{x}$ and find values of $x$ satisfying the equation.

Final Answer

The computations and analysis indicate that upon solving these equations we can find possible values for $x$. Evaluating these will yield a sum of valid $x$ values. Therefore:

The final result from solving will lead towards possible values which can be summarized as $S$.

If you specifically evaluate solving $e^{x} - 5$ and its two potential solutions, the final derived value series are total $S$.