PROOF 126

Prove.

The exponential and logarithmic functions are symmetric about y = x

On real functions.

A way to read this proof

First separate what is assumed from what is to be proved. At each step, ask which definition or earlier result permits the move. If it is unclear, return to that premise rather than skipping to the conclusion.

Definition D1

For a > 0, a ≠ 1, define E(y) = aʸ. The domain is all real numbers.

Proof.

(1)The condition on the base is taken as established in Proof 100.A1

(2)For every real y, aʸ > 0, so the values of E are positive reals.A2

∎

Definition D2

logₐ x means the y satisfying aʸ = x, written L(x) = logₐ x. The domain is the positive reals.

Proof.

So far this is only a name. For the name to be a function, such a y must exist and be unique, which is verified in L1.convention

∎

Lemma L1

E is a one-to-one correspondence from all real numbers onto all positive reals, and for each x > 0 the y with aʸ = x is unique.

Proof.

(1)E(y) > 0 for every y, so all values of E are positive reals.D1, A2

(2)If x > 0, there exists y with aʸ = x.A3

(3)If y ≠ z, then aʸ ≠ aᶻ, so E(y) = E(z) is equivalent to y = z.D1, A4

(4)Therefore the y of (2) is unique, and E is a one-to-one correspondence.(1), (2), (3)

(5)Each x > 0 is assigned exactly one value, so L is a function.(4), D2

∎

Theorem T1

For every x > 0, L(x) = E⁻¹(x). That is, L is the inverse of E, with L(E(y)) = y and E(L(x)) = x.

Proof.

(1)L(x) = y ⟺ aʸ = xD2

(2)aʸ = x ⟺ E(y) = xD1

(3)E is a one-to-one correspondence, so E⁻¹ is defined.L1, A5

(4)E(y) = x ⟺ E⁻¹(x) = y(3), A5

(5)Chaining the three equivalences gives L(x) = E⁻¹(x)(1), (2), (4)

(6)Therefore L(E(y)) = y and E(L(x)) = x.(5), A5

∎

Theorem T2

The two graphs G(E) and G(L) are symmetric about the line y = x.

Proof.

(1)E is a one-to-one correspondence.L1

(2)Therefore G(E) and G(E⁻¹) are symmetric about y = x.(1), A7

(3)L = E⁻¹.T1

(4)Therefore G(E) and G(L) are symmetric about y = x.(2), (3)

∎

Corollary C

The point (0, 1) on the graph of E and the point (1, 0) on the graph of L are reflections of each other about y = x.

Proof.

(1)E(0) = a⁰ = 1, so (0, 1) is an element of G(E).D1, A8

(2)E is a one-to-one correspondence, so (0,1) ∈ G(E) ⟺ (1,0) ∈ G(E⁻¹)L1, A6

(3)L = E⁻¹, so (1, 0) is an element of G(L) and L(1) = 0.(2), T1, D2

(4)Therefore the y-intercept of the exponential function and the x-intercept of the logarithmic function are reflections about y = x.(1), (3), T2

∎

Remark

Can A4 be removed from this proof?

Separate the lines asserting existence from the lines asserting uniqueness. For the logarithm to be a function, both are needed.