Mathematics question - solutions of e^{-x} = ax+b?

Question

The following is a question which was on my younger brother's first-year exam paper for 'Advanced Mathematics for Economists'. I am a mathematics graduate myself and am tutoring him on the syllabus for a retake he is doing in a few weeks. However, I can't get my head around this one question...

Question:

Consider the following equation in x:

e^{-x}=ax+b

where a and b are parameters. Which of the following sufficient condition for a solution to exist?

A. a < 0
B. b = 0
C. a = -1
D. a > 0

Since complex numbers have not been covered in his syllabus, we can assume that any solutions and parameter values will be real.

I am a bit rusty on my rules of exp/log functions... I have got as as ln(x) - x = - ( ln ( ab ) ), which I believe is implied by the above, but still can't see how this helps with a solution. Mathematica doesn't seem to help either...

Answer 1

e^(- x) = ax + b



Consider the two functions y = e^(- x) and y = ax + b



They are equal when the two curves meet.

Sketch a graph of the negative exponential;



it crosses the x axis at y = 1, the gradient increases rapidly for large negative values of x and the x axis is an asymptote for large positive values of x.



Now, how can you place a straight line so that it does or does not cut the curve?



If the gradient is positive (a), regardless of the value of b (the intercept) the line crosses the curve.



This is your answer, the gradient a must be > 0



If the gradient is negative (a < 0) you can always find an intercept (b) so that the line is always below the curve.



If b = 0 you can find a gradient that keeps the line below the curve.



I know that this is not an analytic solution but it helps you focus on what is happening as you vary a & b.

Answer 2

From the bludgeon it to death school of thought comes: (This is not a solution, but some thoughts on the problem, as I don't know how much time I'll have to work on this later. If someone solves it, I'll remove my heavy handed approach). Assume that the final polynomial is x⁶ + Bx + C for B and C rational. One pair of factors must be of the form (x⁴ + bx³ + cx² + dx + e)(x² + gx + h) Since the coefficients on x⁵, x⁴, x³ and x² are all 0, this constrains b and c to be -g and g² - h, respectively. It further constrains d to be 2hg-g³ and e to be g⁴ - 3hg² + h² However, the first factor must actually be the product of two more quadratics: (x² + vx + w) (x² + sx + t) Matching on x⁵ => s = -(g+v) and matching on x⁴ => t = v² + gv + g² - w - h So far we have: (x² + vx + w) (x² - (v+g)x + v² + gv + g² - w - h) (x² + gx + h) Now vt + ws = d or v(v² + gv + g² - h - w) - w(g+v) = 2hg-g³ => w(2v+g) = g³ - 2hg + v(v² +gv + g² - h) = (g² + v²)(g + v) - h(2g+v) However, the messiness rises exponentially (or should I say quadratically?) when we add in that wt = e = g⁴ - 3hg² + h². We get: w(v²+gv+g² - w -h) = e => ((g²+v²)(g+v)-h(2g+v)) (v³ + 2gv² + 2vg² + h(g-v)) = (2v+g)²e => 3h²(g² + gv + v²) + 2h(v⁴ + 2v³g - 3v²g² - 4vg³ - 2g⁴) + (g⁶ + 2g⁵v -5v³g³ - 5g²v⁴ - 3gv⁵ - v⁶) In other words, after solving for the above equation for h in terms of v and g, all the other terms can be expressed in terms of them, so we should get the form x⁶ + Bx + C. The question is whether they are all rational, and that depends on the descriminant. The discriminant is: δ = 4(g+2v)²(g⁶ + 3g⁵v + 3g⁴v² + g³v³ + 3g²v⁴ + 3gv⁵ + v⁶) If v or g is 0, then this is clearly square. The other point at which it is clearly square is when g=-2v. In this case we have: ((g²+v²)(g+v)-h(2g+v)) (v³ + 2gv² + 2vg² + h(g-v)) = 0 or (-5v³ + 3vh) (5v³ - 3vh) = 0 or h = 5v²/3, s = v, t = 4v²/3 - w Thus, we have: (x² + vx + w) (x² + vx + (4v²/3-w)) (x² - 2vx + 5v²/3) = x⁶ + x² ((4v²/3 - w)w + 11v⁴/9) + x(20v⁵/9 - 2v(4v²/3 - w)w) + 5v²(4v²/3 - w)w/3 Given that the x² term must be 0, meaning that (4v²/3 - w)w = -11v⁴/9, this reduces to: x⁶ + x² ((4v²/3 - w)w + 11v⁴/9) + 14xv⁵/3 - 55v⁶/27 So now let's see what w will be: w² - 4wv²/3 - 11v⁴/9 = 0 or 9w² - 12wv² - 11v⁴ = 0 => w = v²( 12 ± √(12² + 4*9*11) ) / (2 * 9) = v²( 2 ± √15 ) / 3, grim news indeed. Therefore, the best that I have been able to come up with (but middle factor is not complex) is: (3x² + 3vx + v²( 2 + √15 )) (3x² + 3vx + v²( 2 - √15 )) (3x² - 6x + 5v²) = 27x⁶ + 126xv⁵ - 55v⁶ The other cases of, say, g=0 are: (x² - v²) (x² + vx + v²) (x² - vx + v²) = x⁶ - v⁶ (3x² + v²) (3x² + 3vx + v²) (3x² - 3vx + v²) = 27x⁶ + v⁶ -------------- In the x⁴+Dx+E case where there are two irreducible, rational, quadratics, they should take the form of x² - x√(R₁+R₂) + R₁ and x² + x√(R₁+R₂) + R₂ with R₁ and R₂ > 0 and R₁+R₂ a square. Thus, E = R₁R₂ and (R₁-R₂)√(R₁+R₂) = D In your example, you used R₁=5 and R₂=4 If you are given E>0 and D, then you could note that (R₁-R₂)²(R₁+R₂) = D² => (R₁+R₂)³ = D² + 4R₁R₂(R₁+R₂) = D² + 4E(R₁+R₂) So, if z³ - 4Ez - D² has a rational root, then you have identified z = R₁+R₂, and hence can determine R₁ and R₂ More importantly, given x² - xS + R where 0 < 4R/3 < S² < 4R, ( x² - xS + R ) ( x² + xS + (S²-R) ) = x⁴ + xS*(2R-S²) + R(S²-R) ------------ Getting back to the original problem, casting one of the complex roots as x=re^(iΘ) means a complimentary root is re^(-iΘ) resulting in a quadratic of (x-re^(iΘ))(x-re^(-iΘ)) = x² - 2rxcos(Θ) + r² = x² - sx + r² where r² and s=rcos(Θ) are rational. Applying this idea to x⁶ + Dx + E = 0 => r⁶e^(6iΘ) + Dre^(iΘ) + E = 0. Separating imaginary and real components means that r⁶sin(6Θ) + Drsin(Θ) = 0 and r⁶ cos(6Θ) + Dr cos(Θ) + E = 0 The first equation means 2cos(Θ) (4cos²(Θ) - 1) (4cos²(Θ) - 3) r⁵ + D = 0 while the second gives (2cos²(Θ)(4cos²(Θ)-3)² - 1)r⁶ + Dr cos(Θ) + E = 0 Using s = 2rcos(Θ) makes these two equations: D = -s(s² - r²)(s² - 3r²) and s²(s²-3r²)² - 2r⁶ + Ds + 2E = 0 = s²(s²-3r²)² - 2r⁶ + -s²(s² - r²)(s² - 3r²) + 2E = s²(s⁴ -6s²r² + 9r⁴) - 2r⁶ + - s²(s⁴-4s²r² + 3r⁴) + 2E => 2E = 2s⁴r² - 6s²r⁴ + 2r⁶ or E = r² (s⁴ - 3s²r² + r⁴) At this point we will revisit the three quadratics. We want: x⁶ + Dx + E = (x² - s₁x + r₁²) (x² - s₂x + r₂²) (x² - s₃x + r₃²) = (x² - s₁x + R₁) (x² - s₂x + R₂) (x² - s₃x + R₃) where R₁ = r₁², R₂ = r₂², R₃ = r₃² and R = R₁ + R₂ + R₃ We have the following relationships: From x⁵: s₁ + s₂ + s₃ = 0 from which also follows that s₁² + s₂² + s₃² = -2 (s₁ s₂ + s₂ s₃ + s₁ s₃) and 3s₁ s₂ s₃ = s₁³ + s₂³ + s₃³ From x⁰: E = R₁ R₂ R₃ From x¹: D = - s₁ R₂ R₃ - R₁ s₂ R₃ - R₁ R₂ s₃ From x²: R₁ R₂ + R₂ R₃ + R₁ R₃ + R₁ s₂ s₃ + s₁ R₂ s₃ + s₁ s₂ R₃ = 0 From x⁴: R₁ + R₂ + R₃ + s₁ s₂ + s₂ s₃ + s₁ s₃ = 0 or R₁ + R₂ + R₃ = s₁² + s₁ s₂ + s₂² From x³: (s₁ R₂ + R₁ s₂) + (s₂ R₃ + R₂ s₃) + (s₁ R₃ + R₁ s₃) + s₁ s₂ s₃ = 0 or s₁ (R₂ + R₃) + s₂ (R₁ + R₃) + s₃ (R₁ + R₂) + s₁ s₂ s₃ = 0 Note that there are 4 constraints (of 0) with 6 unknowns. That leaves two degrees of freedom, just as with the x⁴ + Dx + E case. In other words, if we fix, s₁ and R₁ that will fix all the other unknowns. Of course we have to hope that we can solve the equations and that they will generate rational numbers (coefficients). From the earlier E = r² (s⁴ - 3s²r² + r⁴) we have R₁ R₂ R₃ = R₁ (s₁⁴ - 3s²R₁ + R₁²) or simply R₂ R₃ = s₁⁴ - 3s₁²R₁ + R₁² and from earlier we also have D = -s₁(s₁² - R₁)(s₁² - 3R₁) which is the x¹ coefficient = - s₁ R₂ R₃ - R₁ s₂ R₃ - R₁ R₂ s₃ = - s₁ (s₁⁴ - 3s₁²R₁ + R₁²) - R₁ s₂ R₃ - R₁ R₂ s₃ => s₁ (2R₁² - s₁²R₁) = R₁(s₂ R₃ + R₂ s₃) or s₁ (2R₁ - s₁²) = s₂ R₃ + R₂ s₃ From the x² coefficient:: R₁ R₂ + R₂ R₃ + R₁ R₃ + R₁ s₂ s₃ + s₁ R₂ s₃ + s₁ s₂ R₃ = 0 => R₁ R₂ + R₂ R₃ + R₁ R₃ - R₁ s₂ (s₁ + s₂) - s₁ R₂ (s₁ + s₂) + s₁ s₂ R₃ = 0 => R₁ R₂ + R₂ R₃ + R₁ R₃ = R₂ s₁² + R₁ s₂² + s₁ s₂ (R₁ + R₂ - R₃) => R₁ R₂ + R₂ R₃ + R₁ R₃ = R₂ s₁² + R₁ s₂² + (R - s₁² - s₂²)(R₁ + R₂ - R₃) => R(R₁ + R₂ - R₃) - R₁ R₂ - R₂ R₃ - R₁ R₃ = s₁² (R₁ - R₃) + s₂² (R₂ - R₃) = s₁² (R₁ - R₃) + (R - s₁² - s₁ s₂)(R₂ - R₃) => R₁² - R₂ R₃ = s₁² (R₁ - R₂) - s₁ s₂ (R₂ - R₃) or s₁ s₂ (R₂ - R₃) = s₁² (R₁ - R₂) - (R₁² - R₂ R₃) = s₁² (R₁ - R₂) - (R₁² - (s₁⁴ - 3s₁²R₁ + R₁²)) = s₁² (R₁ - R₂) + s₁⁴ - 3s₁²R₁) = s₁² (s₁² - 2R₁ - R₂) or s₂ (R₂ - R₃) = s₁ (s₁² - 2R₁ - R₂) That is to say, we can express s₂ is a rational function of the other variables: s₂ = s₁ (s₁² - 2R₁ - R₂) / (R₂ - R₃) Thus, we also have s₃ by swapping the 2 and 3 subscripts: s₃ = s₁ (2R₁ + R₃ - s₁²) / (R₂ - R₃) So at this point, everything hinges on finding some way to determine R₂ and R₃. We'll try this by plugging into R₁ + R₂ + R₃ = s₁² + s₁ s₂ + s₂² = s₁² + s₁² (s₁²-2R₁-R₂)²/(R₂ - R₃)² + s₁² (s₁² - 2R₁ - R₂) / (R₂ - R₃) = s₁² (1 + [(s₁²-2R₁-R₂) / (R₂ - R₃) ][ (s₁²-2R₁-R₃) / (R₂ - R₃) ]) or R(R₂ - R₃)² = s₁² [ (R₂ - R₃)² + (s₁²-2R₁-R₂)(s₁²-2R₁-R₃) ] Now let z = R₂ + R₃, and p = R₂ * R₃ = s₁⁴ - 3s₁²R₁ + R₁² => (R₁ + z)(z² - 4p) = s₁² [ z² - 4p + (s₁²-2R₁)² -z(s₁²-2R₁) + p ] or z³ + z²(R₁ - s₁²) + z(s₁⁴ - 2R₁s₁² - 4p) - p(4R₁ - s₁²) - (s₁²-2R₁)² = 0 => z³ + z²(R₁ - s₁²) - z(3s₁⁴ -10R₁s₁² + 4R₁²) + (s₁² - R₁)(2s₁⁴ - 7s₁²R₁ + 4R₁²) = 0 Given rational s₁ and R₁, if we could find a rational solution to the above equation in z (assuming that I haven't made any calculational errors - a big IF - and I suspect I have), we would at least have a fighting chance. However, I am stuck at this point. If I had to guess, I'd think there is no solution. Note although z is solving for the sum R₂ + R₃, it is easy to convert the above to a sixth order polynomial with even powers by using R₃ = E / (R₁ R₂) = (s₁⁴ - 3s₁²R₁ + R₁²) / R₂

Answer 3

It might help to think of the graphs of y = e^(-x) and y = ax + b. The question then becomes: what conditions could we have for a or b that guarantees the two graphs intersect? The first graph looks sort of like this:

http://www.pindling.org/Math/Learning/Graphs/3af2e722.jpg



The second graph is just a line with slope of 'a' and y-intercept 'b'. Notice that the line y = -x (which is just the diagonal line passing through the origin, from the top left to the bottom right, at a 45-degree angle with the x-axis) would never touch the first graph. This means a = -1 and b=0. So neither A, B, nor C are conditions that guarantee the two lines will meet. But "D" works, because as long as the slope is positive, you'll see that the line always touches the other graph.

Answer 4

Look at the graphs of each side of the equation. The graph of e^-x slopes downwards. The graph of ax + b is a straight line, whose gradient and position are determined by a and b.



Case A: If a < 0 the straight line slopes downwards. Just make b very negative and it won't touch the e^-x curve, so this is not sufficient.



Case B: With b = 0, the straight line crosses the origin. Visually you can see, some negative value of a will make sure the lines don't cross.



Case C: As with case A.



Case D: This will clearly always cross the e^-x line as it is sloping upwards.



So D is sufficient.

Answer 5

Well if we take logs of both sides



-x = ln(ax+b)

then ax+b > 0, because we cant log a non-positive number.



Case 1, x is positive

Now, for the result to be negative, ax+b<1.

that means ax<1-b. Nothing.



Case 2, x is negative

ax+b > 0

since x is -ve, b > abs(ax). So b acn only be 0 if a is -ve.



I don't know, question seems dodgy.

Answer 6

e^{-x}=ax+b



If we take the natural log of both sides, it becomes...



-x = ln(ax+b)

x = -ln (ax+b)



But ax+b >0, so...



ax> -b

x> -b/a



Therefore, for there to be a sol'n...



"a" must not be equal to 0 and a<0.



ANSWER: A



Not sure though. i'm quite rusty myself. i'll get back to you when i'm more certain... :P