   Chapter 2.5, Problem 58E

Chapter
Section
Textbook Problem

(a) Prove that the equation has at least one real root. (b) Use your calculator to find an interval of length 0.01 that contains a root.ln x = 3 – 2x

(a)

To determine

To prove: The equation lnx=32x has at least one real root.

Explanation

Theorem used: The Intermediate value Theorem

Suppose that if f is continuous on the closed interval [a, b] and let N be any number between f(a) and f(b), where f(a)f(b). Then there exists a number c in (a, b) such that f(c)=N.

Proof:

Express the equation as, lnx3+2x=0.

Consider the function f(x)=lnx3+2x.

The logarithmic function defined on the interval (0,) and the polynomial function defined on Real numbers.

The function f(x) is a combination of polynomial and logarithmic function and it is defined on the interval (0,). So it is continuous everywhere on its domain (0,).

Without loss of generality, take the sub-interval [1,2].

In order to show that there is at least one root of the equation lnx3+2x=0 in the interval (1,2), it is enough to show that there is a number c between 0 and 1 for which f(c)=0.

Take a=1, b=2 and N=0

(b)

To determine

To find: An interval of length 0.01 that contains a root by using the calculator.

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