(both sides if ;he equation 2 = e”:

I ” 2 = I”&™

(6)

= rt,

˜

,

II... (4) Solvmg (6) for this rule the fact that the doubling time is t = time2)/r.

“˜mg I yields (In

Since In 2 = 0.69, says that to estimaw the doubling for

interest rate I, just divide the interest rate into 69. For example, the doubling

I t”ne at 10 percent interest is 69/1U = 6.9 years; the doubling time at 8 percent

interest is 69/S = 8.625 years. This calculation also tells us that it would take

I 8 625 years for the price level to double if the inflation rate stays wnstant at 8

percent.

As we discussed in Section 3.6, economists studying the relationship between

the price p and the quantity 4 demanded of some good will often choose t”

work with the two-parameter family of constant elasticity demand functions,

CJ = kp”. where k and E are parameters which depend on the good under study.

The parameter F is the most interesting of the two since it equals the elasticity

(p/q)(dq/dp). Taking the log of both sides of 9 = kp” yields:

Inq = Inkp” = Ink + clnp (7)

In logarithmic coordinates, demand is now a linear function whose slope is the

elasticity c.

LOG 93

15.51 DERIVATIVES OF EXPAND

EXERCISES

5.5 Solve the following equations for x:

a) 2”” = 18; b)/ = 1; c)P = 2;

d) 2x-2 = 5./ e)Inx™ = 5: Jyllnx5˜2 -0.5lnx = In25.

5.6 Derive a formula for the amount of time that it takes money to triple in a bank account

that pays interest at rate r compounded continuously.

5.7 How quickly will $500 grow to $600 if the interest rate is 5 percent compounded

continuously?

5.5 DERIVATIVES OF EXP AND LOG

To work effectively with exponential and logarithmic functions, we need to com-

pute and use their derivatives. The natural logarithmic and exponential functions

have particularly simple derivatives, as the statement of the following theorem

indicates.

r

Theorem 5.2 The functions r™ and Inx are continuous functions on their

domains and have continuous derivatives of every order. Their first derivatives

are given by

If u(x) is a differentiable function, then

c) p) = p(d) +.),

d) (I”U(X))™ = s if U(X) > 0.

We will pnwc this theorem in stages. That the exponential map is continuous

should be intuitively clear from the graph in Figure 5.4; its graph has no jumps

or discontinuities. Since the graph of Inx is just the reflection of the graph of 8

across the diagonal {x = y}, the graph of Inn has no discontinuities either, and so

the function lnx is continuous for all x in the set R+_ of positive numbers.

It turns out to be easier to compute the derivative of the natural logarithm first.

94 EXPONENTS AND LOGARITHMS 151

Lemma 5.1 Given that y = lnx is a continuous function on R++, it is also

differentiable and its derivative is given by

(lnx)™ = 5

Proof We start, of course, with the difference quotient that defines the derivative,

and we then simplify it using the basic properties of the logarithm. Fix x > 0.

In(x + h) - Inx +“(˜)=ln(l+y

h

=h(l+$$

1Now, let m = l/h. As h - 0, m - =. Continuing calculation with

OUT

m = l/h, we find

,im In(x + h) - Inn

h

h-O

Therefore, (Inx)™ = l/x. The fact that we can interchange In and lim in the

above string of equalities follows from the fact that y = Inx is a continuous

function: x,,, - sir implies that Ink-,,, - Inxn; or equivalently,

n

l$n(lnx,,,) = In limx,,,

( “/ )

The other three conclusions of Theorem S.2 follow immediately from the

Chain Rule, as we now prove.

If h(x) is a differentiable and positive function, then

Lemma 5.2

!&h(x)) = g

PI-oaf WC simply apply the Chain Rule to the composite function f(x) = In h(x).

The derivative of f is the derivative of the outside function III-which

Iis l/Levdludted at the inside function h(x)-so it™s I/h(x)- times the

15.51 DERl”ATl”ES OF EXPAND LOG 95

of the inside function h:

1 h™(x) .

(Inh(d) = m h™(x) = ho.

We can now easily evaluate the derivative of the exponential function y = PI,

using the fact that it is the inverse of lnx.

(e”)™ = @,

Lemma 5.3

Proof Use the definition of Inx in (4) to write In e™ = x. Taking the derivative

of both sides of this equation and using the previous lemma, we compute

It follows that

($)™ = r*, .

Finally, to prove part c of Theorem 5.2, we simply apply the Chain Rule to the