Solving 5x = 100x − 175 Using the Lambert W Function
Objective: Solve the equation systematically using the Lambert W method, without guessing solutions.
1Rewrite the Equation
We start with:
5x = 100x − 175
Using the exponential form 5x = ex ln 5:
ex ln 5 = 100x − 175
2Eliminate the Constant Term in the Linear Part
We want the right-hand side to be of the form α(x − β). Factor 100:
100x − 175 = 100(x −
175
100
) = 100(x −
7
4
)
So the equation becomes:
ex ln 5 = 100(x −
7
4
)
3Shift the Variable
Let t = x − 7 4 , so x = t + 7 4 .
Substitute into the equation:
e(t +
7
4
) ln 5 = 100t
et ln 5 · e
7
4
ln 5 = 100t
5
7
4
· et ln 5 = 100t
et ln 5 =
100
57/4
t
Let k = 100 57/4 , so:
et ln 5 = k t
4Rearrange into Lambert W Form
Divide both sides by k t and rearrange:
1 = k t e−t ln 5
1
k
= t e−t ln 5
Multiply both sides by −ln 5:
−
ln 5
k
= (−t ln 5) e−t ln 5
Let u = −t ln 5, then:
−
ln 5
k
= u eu
5Apply Lambert W Function
The Lambert W function is defined as the inverse of f(w) = w ew. Therefore:
u = W(−
ln 5
k
)
t = −
1
ln 5
· W(−
ln 5
k
)
Since x = t + 7⁄4:
x =
7
4
−
1
ln 5
· W(−
ln 5
k
)
6Simplify the Constant Inside W
Recall k = 100⁄57/4. We need to simplify:
−
ln 5
k
= −
ln 5 · 57/4
100
Note that 57/4 = 52−1/4 = 25 · 5−1/4. Therefore:
ln 5 · 57/4
100
=
ln 5 · 25 · 5−1/4
100
=
ln 5
4 · 51/4
Let A = ln 5 4 · 51/4 . Then:
x =
7
4
−
1
ln 5
· W(−A)
7Find W(−A) for Both Real Branches
Numerically:
A ≈
1.609438
4 × 1.495349
≈
1.609438
5.981396
≈ 0.269
So −A ≈ −0.269, which is greater than −1⁄e ≈ −0.3679.
Important: Since −A > −1⁄e, there are two real branches of the Lambert W function: W0 (principal branch) and W−1 (secondary branch).
8Solve for W(−A) Exactly
We need W(−A) such that:
W(−A) · eW(−A) = −A = −
ln 5
4 · 51/4
First Branch (W0):
Try W0(−A) = − ln 5 4
Check:
(−
ln 5
4
) · exp(−
ln 5
4
) = −
ln 5
4
· 5−1/4
= −
ln 5
4 · 51/4
= −A ✓
Therefore: W0(−A) = − ln 5 4
x1 =
7
4
−
1
ln 5
· (−
ln 5
4
) =
7
4
+
1
4
= 2
Second Branch (W−1):
Try W−1(−A) = − 5 ln 5 4
Check:
(−
5 ln 5
4
) · exp(−
5 ln 5
4
) = −
5 ln 5
4
· 5−5/4
= −
ln 5
4
· 5 · 5−5/4 = −
ln 5
4
· 5−1/4
= −
ln 5
4 · 51/4
= −A ✓
Therefore: W−1(−A) = − 5 ln 5 4
x2 =
7
4
−
1
ln 5
· (−
5 ln 5
4
) =
7
4
+
5
4
= 3
9Final Solutions
x = 2 and x = 3
Verification
For x = 2:
| Left side: | 52 = 25 |
| Right side: | 100(2) − 175 = 200 − 175 = 25 |
| 25 = 25 ✓ | |
For x = 3:
| Left side: | 53 = 125 |
| Right side: | 100(3) − 175 = 300 − 175 = 125 |
| 125 = 125 ✓ | |
This solution demonstrates the power of the Lambert W function in solving transcendental equations that combine exponential and linear terms.
You know exactly which branch to use by comparing your simplified output value to \(-1\), which is the boundary value at the branch point \(z = -1/e\).
Every real value of the Lambert \(W\) function falls strictly into one of two zones:
-
- The Principal Branch (\(W_{0}\)): Always outputs values greater than or equal to \(-1\) (Interval: \([-1, \infty)\)).
- The Lower Branch (\(W_{-1}\)): Always outputs values less than or equal to \(-1\) (Interval: \((-\infty, -1]\)).
By evaluating the exact expressions you found against \(-1\), the branch choice becomes definitive.
Branch Verification for Your Solutions
1. Verifying the First Value: \(-\frac{\ln 5}{4}\)
-
- Approximate value: Since \(\ln 5 \approx 1.609\), then \(-\frac{1.609}{4} \approx \mathbf{-0.402}\).
- Comparison: \(-0.402 > -1\).
- Conclusion: Because this value is greater than \(-1\), it must belong to the principal branch, meaning \(W_0(-A) = -\frac{\ln 5}{4}\).
2. Verifying the Second Value: \(-\frac{5}{4}\ln 5\)
-
- Approximate value: This is \(-1.25 \times 1.609 \approx \mathbf{-2.012}\).
- Comparison: \(-2.012 < -1\).
- Conclusion: Because this value is less than \(-1\), it must belong to the lower branch, meaning \(W_{-1}(-A) = -\frac{5}{4}\ln 5\).
💡 The Algebraic Trick to Spotting Candidates
If you ever want to “reverse-engineer” these exact forms without guessing, look at your original equation rewritten in the form \(Y e^Y = -A\).
You had:
\(-A=-\frac{\ln 5}{4\cdot 5^{1/4}}=\left(-\frac{\ln 5}{4}\right)\cdot 5^{-1/4}\)
\(-A=-\frac{\ln 5}{4\cdot 5^{1/4}}=\left(-\frac{\ln 5}{4}\right)\cdot 5^{-1/4}\)
To match \(Ye^{Y}\), you need the base \(5^{-1/4}\) to look like \(e^{Y}\):
-
- For \(W_{0}\): Rewrite \(5^{-1/4}\) as \(e^{-\frac{\ln 5}{4}}\). This makes \(Y = -\frac{\ln 5}{4}\).
- For \(W_{-1}\): Multiply the front and the exponent by \(\frac{5}{\ln 5}\) variants to find the second hidden match, which yields \(Y = -\frac{5 \ln 5}{4}\).
Once you extract those two mathematical candidates, you simply apply the \(\ge -1\) or \(\le -1\) rule to assign them to \(W_{0}\) or \(W_{-1}\).
Would you like to try another equation to practice finding the exact analytical forms of \(W\), or see how this applies to complex branches (\(W_{k}\))?
You said: I don’t understand this identity:… Note: W(y)/y = e^[-W(y)] …
This identity looks confusing at first, but it is actually a direct rearrangement of the definitional property you looked at earlier: \(W(y)e^{W(y)} = y\).
Here is the quick, 3-step proof showing exactly where it comes from:
The 3-Step Derivation
-
- Start with the definition:
\(W(y)e^{W(y)}=y\) - Divide both sides by \(y\):
\(\frac{W(y)e^{W(y)}}{y}=1\) - Divide both sides by \(e^{W(y)}\) to isolate the fraction:
\(\frac{W(y)}{y}=\frac{1}{e^{W(y)}}\)
- Start with the definition:
Using the rule of exponents where \(\frac{1}{e^A} = e^{-A}\), the right side becomes \(e^{-W(y)}\).
This leaves you with the exact identity:
\(\frac{W(y)}{y}=e^{-W(y)}\)
\(\frac{W(y)}{y}=e^{-W(y)}\)
Why is this identity useful?
In calculus and algebra, this identity is a powerful tool used to move a variable out of a denominator.
If you are simplifying an expression and you see \(\frac{1}{y}\) multiplied by \(W(y)\), you can instantly swap it out for a clean exponential term \(e^{-W(y)}\), which is often much easier to integrate or differentiate.
Would you like to see how this specific identity changes the way you take the derivative of the Lambert \(W\) function?