Solving \( 5^x = 100x – 175 \) Exactly Using the Lambert W Function — No Guesswork
Most AI solvers can handle algebraic equations, but they often fail with mixed exponential-linear forms such as:
\[ 5^x = 100x – 175 \]
In this article, we’ll derive exact symbolic solutions using the Lambert W function — without guessing, rounding, or numerical approximations.
🧮 Step 1: Rearrange the Equation
Start from: \[ 5^x = 100x – 175 \] Move constants and simplify: \[ 5^x = 100(x – 1.75) \] Divide both sides by \(100\): \[ \frac{5^x}{100} = x – 1.75 \] Let \( y = x – 1.75 \Rightarrow x = y + 1.75 \). Substituting: \[ \frac{5^{y + 1.75}}{100} = y \]
⚙️ Step 2: Prepare for the Lambert W Form
Rewrite the equation: \[ y e^{-y \ln 5} = \frac{5^{1.75}}{100} \] Multiply both sides by \(-\ln 5\): \[ (-y\ln 5)e^{-y\ln 5} = -\frac{\ln 5}{100}5^{1.75} \] Let \( u = -y\ln 5 \). Then the equation becomes: \[ u e^{u} = -\frac{\ln 5}{100}5^{1.75} \] By definition of the Lambert W function: \[ u = W_k\!\left(-\frac{\ln 5}{100}5^{1.75}\right) \]
🧩 Step 3: Solve for \(x\)
\[ -y\ln 5 = W_k\!\left(-\frac{\ln 5}{100}5^{1.75}\right) \Rightarrow y = -\frac{1}{\ln 5}W_k\!\left(-\frac{\ln 5}{100}5^{1.75}\right) \] \[ x = 1.75 – \frac{1}{\ln 5}W_k\!\left(-\frac{\ln 5}{100}5^{1.75}\right) \]
🌿 Step 4: Understanding the Two Real Branches
The argument of \(W\) lies in the range \((-1/e, 0)\), so both real branches \(W_0\) and \(W_{-1}\) exist.
\[ x = 1.75 – \frac{1}{\ln 5}W_k\!\left(-\frac{\ln 5}{100}5^{1.75}\right), \quad k \in \{0, -1\} \]
| Branch | Range | Exact \(x\) | Interpretation |
|---|---|---|---|
| \(W_0\) | \(W_0(z) \ge -1\) | \(x = 2\) | Principal branch |
| \(W_{-1}\) | \(W_{-1}(z) \le -1\) | \(x = 3\) | Lower branch |
Thus, the two exact integer solutions are: \[ \boxed{x = 2 \text{ (from } W_0)}, \quad \boxed{x = 3 \text{ (from } W_{-1})} \]
🔬 Step 5: Why Many AI Solvers Miss This
- They fail to isolate \(x\) into the canonical form \(u e^u = k\).
- They ignore the domain restriction \(-1/e \le z < 0\).
- They don’t consider both real branches of the Lambert W function.
That’s why numeric approximations appear, while the analytic Lambert W method yields two exact integer roots.
📊 Step 6: Comparison Table
| Method | Result | Explanation |
|---|---|---|
| DeepSeek (AI) | Approximate roots only | Stopped before canonical W-form reduction |
| Analytical (Lambert W) | \(x = 2, 3\) | Exact symbolic solutions from \(W_0\) and \(W_{-1}\) |
🚀 Conclusion
The Lambert W function provides a precise and elegant way to solve transcendental equations involving exponentials and linear terms. Here, it revealed two exact integer solutions where even powerful AI systems relied on approximations.
\[ x = \frac{7}{4} – \frac{1}{\ln 5}W_k\!\left(-\frac{\ln 5}{100}5^{7/4}\right), \quad k \in \{0, -1\} \]
This corresponds exactly to \(x = 2\) and \(x = 3\).
🧭 Tags
Lambert W function, algebra, transcendental equations, exponential equations, symbolic math, AI math, DeepSeek, advanced algebra