Solve \( 5^x = 100x – 175 \)
Alright — let’s solve \( 5^x = 100x – 175 \) from scratch using the Lambert W method systematically, without guessing solutions.
—
1. Rewrite the equation
We start with:
\[
5^x = 100x – 175
\]
\[
e^{x \ln 5} = 100x – 175
\]
—
2. Eliminate the constant term in the linear part
We want the right-hand side to be of the form \( \alpha (x – \beta) \).
Factor 100:
\[
100x – 175 = 100\left(x – \frac{175}{100}\right) = 100\left(x – \frac{7}{4}\right)
\]
So the equation becomes:
\[
e^{x \ln 5} = 100\left(x – \frac{7}{4}\right)
\]
—
3. Shift the variable
Let \( t = x – \frac{7}{4} \), so \( x = t + \frac{7}{4} \).
Substitute:
\[
e^{\left(t + \frac{7}{4}\right) \ln 5} = 100 t
\]
\[
e^{t \ln 5} \cdot e^{\frac{7}{4} \ln 5} = 100 t
\]
\[
5^{7/4} \cdot e^{t \ln 5} = 100 t
\]
\[
e^{t \ln 5} = \frac{100}{5^{7/4}} \, t
\]
Let \( k = \frac{100}{5^{7/4}} \), so:
\[
e^{t \ln 5} = k \, t
\]
—
4. Rearrange into Lambert W form
\[
1 = k \, t \, e^{-t \ln 5}
\]
\[
\frac{1}{k} = t \, e^{-t \ln 5}
\]
Multiply both sides by \(-\ln 5\):
\[
-\frac{\ln 5}{k} = (-t \ln 5) \, e^{-t \ln 5}
\]
Let \( u = -t \ln 5 \), then:
\[
-\frac{\ln 5}{k} = u \, e^{u}
\]
—
5. Apply Lambert W
\[
u = W\left( -\frac{\ln 5}{k} \right)
\]
\[
t = -\frac{1}{\ln 5} \, W\left( -\frac{\ln 5}{k} \right)
\]
\[
x = t + \frac{7}{4} = \frac{7}{4} – \frac{1}{\ln 5} \, W\left( -\frac{\ln 5}{k} \right)
\]
Recall \( k = \frac{100}{5^{7/4}} \), so:
\[
-\frac{\ln 5}{k} = -\frac{\ln 5}{\frac{100}{5^{7/4}}} = -\frac{\ln 5 \cdot 5^{7/4}}{100}
\]
Thus:
\[
x = \frac{7}{4} – \frac{1}{\ln 5} \, W\left( -\frac{\ln 5 \cdot 5^{7/4}}{100} \right)
\]
—
6. Simplify the constant inside \( W \)
\[
5^{7/4} = 5^{1 + 3/4} = 5 \cdot 5^{3/4}
\]
But better: \( 5^{7/4} = 25 \cdot 5^{-1/4} \) because \( 5^{7/4} = 5^{2 – 1/4} = 25 \cdot 5^{-1/4} \).
Check: \( 2 – 1/4 = 7/4 \), yes.
So:
\[
\frac{\ln 5 \cdot 5^{7/4}}{100} = \frac{\ln 5 \cdot 25 \cdot 5^{-1/4}}{100} = \frac{\ln 5}{4} \cdot 5^{-1/4}
\]
Thus:
\[
-\frac{\ln 5 \cdot 5^{7/4}}{100} = -\frac{\ln 5}{4 \cdot 5^{1/4}}
\]
Let \( A = \frac{\ln 5}{4 \cdot 5^{1/4}} \).
Then:
\[
x = \frac{7}{4} – \frac{1}{\ln 5} \, W(-A)
\]
—
7. Find \( W(-A) \) for both real branches
We know \( W(z) \) is real for \( z \ge -1/e \).
Compute \( A \):
\[
A = \frac{\ln 5}{4 \cdot 5^{1/4}}
\]
Numerically: \( \ln 5 \approx 1.609438 \), \( 5^{1/4} \approx 1.495349 \), so:
\[
A \approx \frac{1.609438}{4 \times 1.495349} \approx \frac{1.609438}{5.981396} \approx 0.269
\]
So \( -A \approx -0.269 \), which is \( > -1/e \approx -0.3679 \), so two real branches \( W_0 \) and \( W_{-1} \).
—
8. Solve for \( W(-A) \) exactly
We want \( W(-A) \) such that:
\[
W(-A) e^{W(-A)} = -A = -\frac{\ln 5}{4 \cdot 5^{1/4}}
\]
First branch:
Try \( W_0(-A) = -\frac{\ln 5}{4} \).
Check:
\[
\left(-\frac{\ln 5}{4}\right) \exp\left(-\frac{\ln 5}{4}\right) = -\frac{\ln 5}{4} \cdot 5^{-1/4} = -\frac{\ln 5}{4 \cdot 5^{1/4}} = -A
\]
✓ Exact.
So:
\[
W_0(-A) = -\frac{\ln 5}{4}
\]
\[
x_1 = \frac{7}{4} – \frac{1}{\ln 5} \left( -\frac{\ln 5}{4} \right) = \frac{7}{4} + \frac{1}{4} = 2
\]
—
Second branch:
Try \( W_{-1}(-A) = -\frac{5}{4} \ln 5 \).
Check:
\[
\left(-\frac{5}{4} \ln 5\right) \exp\left(-\frac{5}{4} \ln 5\right) = -\frac{5}{4} \ln 5 \cdot 5^{-5/4}
\]
\[
= -\frac{\ln 5}{4} \cdot 5^{1} \cdot 5^{-5/4} = -\frac{\ln 5}{4} \cdot 5^{-1/4} = -\frac{\ln 5}{4 \cdot 5^{1/4}} = -A
\]
✓ Exact.
So:
\[
W_{-1}(-A) = -\frac{5}{4} \ln 5
\]
\[
x_2 = \frac{7}{4} – \frac{1}{\ln 5} \left( -\frac{5}{4} \ln 5 \right) = \frac{7}{4} + \frac{5}{4} = 3
\]
—
9. Final solutions
\[
x = 2 \quad \text{and} \quad x = 3
\]
—
\[
\boxed{2, 3}
\]