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Worked solution

Find a logarithmic tangent and a two-intersection slope range

The curve CC has equation 2−y=x2^{-y}=x. Its point AA has x-coordinate a>0a>0. Show that dy/dx=−1/(aln⁡2)dy/dx=-1/(a\ln2) at AA, and find the equation of the tangent there.

Hence find the tangent to CC that passes through the origin. The line y=mxy=mx intersects CC at two distinct points. Write down the range of mm.

[2008/RJC/I/11(a)]

CJC · Discussion · Question 8

Answer

Tangent at AA: y=−x/(aln⁡2)+(1−ln⁡a)/ln⁡2y=-x/(a\ln2)+(1-\ln a)/\ln2. Through the origin: y=−x/(eln⁡2)y=-x/(e\ln2). Two intersections: −1/(eln⁡2)<m<0-1/(e\ln2)<m<0.

Differentiate and write the tangent at A

Taking logarithms gives −yln⁡2=ln⁡x-y\ln2=\ln x, so y=−ln⁡x/ln⁡2y=-\ln x/\ln2 and y′=−1/(xln⁡2)y^{\prime}=-1/(x\ln2). At A=(a,−ln⁡a/ln⁡2)A=(a,-\ln a/\ln2) the gradient is −1/(aln⁡2)-1/(a\ln2), as required. The tangent is

y+ln⁡aln⁡2=−x−aaln⁡2,y=−xaln⁡2+1−ln⁡aln⁡2.y+\frac{\ln a}{\ln2}=-\frac{x-a}{a\ln2},\qquad y=-\frac{x}{a\ln2}+\frac{1-\ln a}{\ln2}.

How did this part go?

Require the tangent to pass through the origin

At (0,0)(0,0), 1−ln⁡a=01-\ln a=0, so a=ea=e. The required line is y=−x/(eln⁡2)y=-x/(e\ln2); it touches the curve at (e,−1/ln⁡2)(e,-1/\ln2). The source calculator plot shows this tangent and the logarithmic curve.

How did this part go?

Find the strict slope interval

The printed graph gives −1/(eln⁡2)<m<0-1/(e\ln2)<m<0. To check the number of intersections, write m=−ln⁡x/(xln⁡2)m=-\ln x/(x\ln2). This ratio decreases to its unique minimum −1/(eln⁡2)-1/(e\ln2) at x=ex=e, then increases towards 00 from below. Thus precisely the slopes strictly between that minimum and 00 give two distinct intersections, both with x>1x>1. At the minimum there is only one tangent contact, and at m=0m=0 there is only the intersection (1,0)(1,0).

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Compare your sketch

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Exact logarithmic curve y=−ln(x)/ln(2), x>0, crossing (1,0), together with its through-origin tangent y=−x/(e ln(2)). They touch at x=e, y=−1/ln(2); the curve is above the tangent elsewhere.
Exact logarithmic curve y=−ln(x)/ln(2), x>0, crossing (1,0), together with its through-origin tangent y=−x/(e ln(2)). They touch at x=e, y=−1/ln(2); the curve is above the tangent elsewhere.
Exact logarithmic curve y=−ln(x)/ln(2), x>0, crossing (1,0), together with its through-origin tangent y=−x/(e ln(2)). They touch at x=e, y=−1/ln(2); the curve is above the tangent elsewhere.

Common mistakes

The exponential is 2 to the power −y, not 2 minus y.

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