x times 2 to the power x equals 8 when x is 2. Change the 8 to 10 and logs alone will not isolate x. So mathematicians gave the answer a name.
Name ____________________Date ____________
Tonight's idea
You can spot the answer to \(x \cdot 2^x = 8\): it is \(x = 2\). Now change 8 to 10. Taking logs gives \(\log_2 x + x = \log_2 10\), and \(x\) is still in two places, once inside a log and once outside. Logs alone cannot isolate \(x\).
Yet there is an answer. For \(x \gt 0\) the quantity \(x \cdot 2^x\) climbs steadily from near 0, with no jumps and no ceiling, so it passes 10 exactly once.
When an answer exists but the usual tools will not isolate it, mathematicians name it. You know one case: \(\sqrt2\) names the positive number whose square is 2.
Lambert W. For \(z \ge 0\), \(W(z)\) is the one number \(y \ge 0\) with \(y e^y = z\). So \(W\) undoes \(y e^y\), as \(\ln\) undoes \(e^y\).
\(W\) is one of many special functions: functions that earned a name because problems kept needing them. It is named after Johann Lambert, who worked in the 1700s. Few calculators have a \(W\) key, but many computer math(s) systems do.
Watch one: trap the answer by halving
Solve \(x \cdot 2^x = 10\). At \(x = 2\) the left side is 8, too small. At \(x = 3\) it is 24, too big. Try the midpoint, see which half holds the answer, and repeat. This is bisection.
step
try \(x\)
\(x \cdot 2^x\)
verdict
so the answer is between
1
2.5
14.14
too big
2 and 2.5
2
2.25
10.70
too big
2 and 2.25
3
2.125
9.27
too small
2.125 and 2.25
4
2.1875
9.96
too small
2.1875 and 2.25
5
2.21875
10.33
too big
2.1875 and 2.21875
Every number between 2.1875 and 2.21875 rounds to 2.2, so \(x = 2.2\) to one decimal place. Each step halves the trap: it started 1 wide, so after 5 steps it is \(1/32\) wide. Bisection is slow, but it cannot fail here: the left side has no jumps, so it cannot get from below 10 to above 10 without passing 10.
Your turn
Carry on the worked example. (a) Do steps 6, 7 and 8. (b) What is \(x\) to two decimal places, and how do you know? (c) The trap starts 1 wide and halves at every step. How many steps, counted from the start, make it narrower than 0.001?
Now name the answer. (a) Explain why \(2^x = e^{x \ln 2}\). (b) Put \(y = x \ln 2\) and show that \(x \cdot 2^x = 10\) turns into \(y e^y = 10 \ln 2\). (c) So \(x = W(10 \ln 2) / \ln 2\). A computer gives \(W(6.9315) = 1.5184\) to four places. Find \(x\) and compare it with exercise 1.
The curve \(z = y e^y\). Its lowest point is at \(y = -1\).The figure shows \(z = y e^y\). Calculus shows that its lowest point is at \(y = -1\), where \(z = -1/e \approx -0.368\), and that the curve falls all the way to that point and rises ever after. (a) Explain, without the figure, why \(y e^y = z\) has exactly one solution when \(z \gt 0\). Treat \(y \le 0\) and \(y \gt 0\) separately. (b) Check that \(y = -\ln 2\) and \(y = -2 \ln 2\) both give \(y e^y = -\tfrac12 \ln 2\). (c) Use the figure to say how many solutions \(y e^y = z\) has when \(z = -0.2\), when \(z = -1/e\) and when \(z = -1\).
Solve \(x^x = 100\) for \(x \gt 1\). (a) Show that \(x\) is between 3 and 4. (b) Find \(x\) to one decimal place by trial. (c) Take \(\ln\) of both sides, put \(u = \ln x\), and show that \(u e^u = \ln 100\), so that \(x = e^{W(\ln 100)}\). A computer gives \(W(4.6052) = 1.2802\). Find \(x\) to two decimal places.
Stretch: a tower of powers
Start with \(a_1 = \sqrt2\) and make each new term by \(a_{n+1} = (\sqrt2)^{a_n}\). Each term puts one more \(\sqrt2\) under the tower \(\sqrt2^{\sqrt2^{\sqrt2}}\) and so on. (a) Compute the first five terms. (b) Show that every term is bigger than the one before, and that every term is below 2. A sequence that keeps rising but stays below 2 must settle on a limit \(T \le 2\) (a fact you may use), and then \((\sqrt2)^T = T\). (c) Check that \(T = 2\) and \(T = 4\) both satisfy \((\sqrt2)^T = T\). Put \(y = -\tfrac12 T \ln 2\) and use exercise 3 to show there are no other solutions. Which one is the tower?
Answers
(a) Step 6: \(x = 2.203125\) gives 10.14, too big. Step 7: \(x = 2.1953125\) gives 10.05, too big. Step 8: \(x = 2.19140625\) gives 10.01, too big. The answer is between 2.1875 and 2.19140625. (b) Every number in that trap rounds to 2.19, so \(x = 2.19\) to two decimal places. After step 7 the trap still reached 2.1953, which rounds to 2.20, so step 8 was needed. (c) After \(n\) steps the width is \(1/2^n\). Since \(2^9 = 512\) and \(2^{10} = 1024\), it takes 10 steps.
(a) \(2 = e^{\ln 2}\), so \(2^x = (e^{\ln 2})^x = e^{x \ln 2}\). (b) The equation is \(x e^{x \ln 2} = 10\). Multiply both sides by \(\ln 2\): \((x \ln 2) e^{x \ln 2} = 10 \ln 2\), which is \(y e^y = 10 \ln 2\). (c) \(10 \ln 2 = 6.9315\) is positive, so \(y = W(6.9315) = 1.5184\) is the only solution, and \(x = 1.5184 / \ln 2 = 2.1906\) to four places. That sits inside the trap from exercise 1 and rounds to 2.19.
(a) If \(y \le 0\) then \(y e^y \le 0\), because \(e^y\) is always positive. So no such \(y\) works. For \(y \gt 0\), both \(y\) and \(e^y\) are positive and increasing, so the product \(y e^y\) is increasing: it cannot take the same value twice. It starts at 0 when \(y = 0\), has no jumps and grows past every number, so it takes the value \(z\) exactly once. (b) \(e^{-\ln 2} = \tfrac12\), so the first gives \(-\tfrac12 \ln 2\). \(e^{-2 \ln 2} = \tfrac14\), so the second gives \(-2 \ln 2 \times \tfrac14 = -\tfrac12 \ln 2\), about \(-0.347\). (c) Two for \(z = -0.2\) (near \(y = -2.54\) and \(y = -0.26\)), one for \(z = -1/e\) (only \(y = -1\)), none for \(z = -1\), which is below the lowest point. In general there are two solutions when \(-1/e \lt z \lt 0\), so there \(W\) has two branches and you must say which one you mean.
(a) \(3^3 = 27\) is too small and \(4^4 = 256\) is too big. For \(x \gt 1\), \(x^x\) is increasing and has no jumps, so there is exactly one solution and it is between 3 and 4. (b) \(3.5^{3.5} = 80.2\), \(3.6^{3.6} = 100.6\) and \(3.55^{3.55} = 89.8\). So \(x\) is between 3.55 and 3.6, and \(x = 3.6\) to one decimal place. (c) \(\ln(x^x) = x \ln x\), so \(x \ln x = \ln 100\). With \(u = \ln x\) you have \(x = e^u\), so \(u e^u = \ln 100\). Since \(\ln 100 = 4.6052\) is positive, \(u = W(\ln 100) = 1.2802\) and \(x = e^{1.2802} = 3.60\) to two places (more exactly 3.597).
Stretch. (a) 1.4142, 1.6325, 1.7608, 1.8409, 1.8927. (b) Because \(\sqrt2 \gt 1\), a bigger exponent gives a bigger power. First, \(a_2 = \sqrt2^{\sqrt2} \gt \sqrt2^{1} = a_1\). Then if \(a_n \gt a_{n-1}\), raising \(\sqrt2\) to each side gives \(a_{n+1} \gt a_n\), so the rise carries on for ever. In the same way \(a_1 \lt 2\), and if \(a_n \lt 2\) then \(a_{n+1}\) is below \(\sqrt2^{2} = 2\) as well. (c) \(\sqrt2^{2} = 2\) and \(\sqrt2^{4} = 4\). Write \(\sqrt2^{T} = e^{\frac12 T \ln 2} = e^{-y}\). Then \(e^{-y} = T\) gives \(T e^{y} = 1\), and multiplying by \(-\tfrac12 \ln 2\) gives \(y e^y = -\tfrac12 \ln 2\). That value is between \(-1/e\) and 0, so by exercise 3 there are exactly two solutions, \(y = -\ln 2\) and \(y = -2\ln 2\), which are \(T = 2\) and \(T = 4\). The limit is at most 2, so the tower is 2. The two branches of \(W\) are the two candidates, and only one of them is the tower.