Hands first, words secondDrag the point P along the solid curve and watch how its mirror image Q moves along the dashed one. Then lower the base a slowly from 2 and keep an eye on the number of “common points”.
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Try this
Drag P. How are the coordinates of P related to those of Q? How does the segment PQ sit relative to the line y=x?
Lower a from 2. At roughly what value of a do the two curves start to have common points? (Try a=1.44.)
Go to 0<a<1, say a=0.5. Where are the common points? Then try a=0.0625.
Switch to “Power functions” and click through the values of α. Which point do all the graphs pass through? How does the monotonicity on (0,+∞) differ between α>0 and α<0?
y=ax (a>0, a=1) has domain R and range (0,+∞); its inverse is y=logax, with domain (0,+∞) and range R. Both are increasing when a>1 and both decreasing when 0<a<1. The graphs of two mutually inverse functions are mirror images in the line y=x:
(t,at)lies ony=ax⟺(at,t)lies ony=logax.
The graph of every power functiony=xα (α a constant) passes through (1,1); on (0,+∞) it is increasing if α>0 and decreasing if α<0. For α=qp (in lowest terms, q>0): if q is even, the domain contains no negative numbers; if q is odd, the function is even when p is even and odd when p is odd.
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Click all eight values of α and use the three readings on the right to check the paragraph above.
Predict first
How many common points do the graphs of y=1.2x and y=log1.2x have?
Choose an answer first; then you can see the result.
✓You predicted it!✗Not what you expected — which is exactly why it is worth a pause.
Many people draw from memory “one curve above y=x, one below” and answer 0. That is what happens for a=2 or a=10.
When the base is close to 1, y=1.2x grows very slowly, so it first crosses the line y=x and then crosses back: 1.2x=x has two solutions (about 1.26 and 14.77). Mirror them, and the log curve meets the exponential curve at the same two points.
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The second common point is at x≈14.77, outside the frame, but it is in the readout. Increase a slowly: when do the two common points merge into one?
The borderline base is a=e1/e≈1.4447: there the curve y=ax just touches y=x at (e,e).
Why: for a > 1 the common points all lie on the mirror
For a > 1, every common point of y = aˣ and y = logₐx lies on the line y = x
Step 1 / 5
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The mirror relation
P is on the exponential curve; swapping its coordinates gives Q on the log curve, and the line y=x is the perpendicular bisector of PQ.
y=ax⟺x=logay
Name a common point
Use monotonicity
A bigger base never touches
The proof breaks for 0 < a < 1
Remove one condition
Break it
Counterexample lab: remove one condition
Conclusion: every common point of y=ax and y=logax lies on the line y=x← still true?
a=1.2: both common points land on the dashed line y=x.
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Change the base to a=161=0.0625 and read off the number of common points.
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There are 3! Besides the one on y=x, there are (41,21) and (21,41): (161)1/4=21 and (161)1/2=41. These two points are mirror images of each other in y=x, and neither lies on the mirror. A decreasing function and its inverse can meet “crosswise”.
Practice
Exercise 1Basic
Solve 2x=8x−1.
Hint · direction
Write both sides with the same base.
Hint · method
8=23. An exponential function is an injection, so equal powers of the same base have equal exponents.
Hint · first step
2x=23(x−1), so x=3x−3.
Full solution
x=23.
Exercise 2Basic
Find the maximum of y=log21(x2−4x+5).
Hint · direction
Is the outer log21 increasing or decreasing?
Hint · method
The outer function is decreasing: the smaller the argument, the larger y. Find the minimum of the argument first.
Hint · first step
x2−4x+5=(x−2)2+1≥1.
Full solution
The argument is at least 1 (equal at x=2), so the maximum of y is log211=0.
Exercise 3Advanced
The interval on which y=log21(x2−2x−3) is increasing is (−∞,t). Find t.
Hint · direction
Find the domain first.
Hint · method
The domain is x<−1 or x>3. The outer function is decreasing, so y increases exactly where the inner function decreases.
Hint · first step
x2−2x−3 is decreasing on (−∞,1); intersect this with the domain.
Full solution
y is increasing on (−∞,−1), so t=−1.
Exercise 4Challenge
How many real solutions does (161)x=log161x have?
Hint · direction
First find the solution on y=x, then think about whether there are solutions that come in “mirror pairs”.
Hint · method
If x0 is a solution, let y0=(161)x0; then x0=(161)y0. Try x0=41 and 21.
Hint · first step
(161)1/4=21 and log16141=21, so x=41 is a solution; in the same way x=21 is one too.
Full solution
Also, (161)x=x has one more solution in (41,21) (the left side decreases, the right side increases, and their order is reversed at the two ends), and it satisfies the original equation too. That makes 3 real solutions. You can see them in the mirror picture with a=0.0625.
Finding where a composite log function increases, forgetting the domain
✗
The wrong step: y=log21(x2−2x−3): the inner x2−2x−3 is decreasing on (−∞,1) and the outer function is decreasing, so y is increasing on (−∞,1).
What went wrong? Think first, then open
DiagnosisThe argument of a logarithm must be positive: x2−2x−3>0, that is, x<−1 or x>3. Points of [−1,1) are not in the domain at all; at x=0, for instance, the argument is −3.
✓
The right step: Find the domain first, then study the inner function's monotonicity inside it: y is increasing on (−∞,−1) and decreasing on (3,+∞).
“With a logarithm, settle the argument first, then talk about monotonicity”
Power, Exponential and Logarithmic Functions ← → step · F full screen · Esc exit