# The equation, ln ( x 2 − 1 ) = 3 for x . ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805 ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 1.6, Problem 50E

(a)

To determine

## To solve: The equation, ln(x2−1)=3 for x.

Expert Solution

The solutions for the given equation are x=±e3+1.

### Explanation of Solution

The given equation is ln(x21)=3.

Take exponential on both sides of the above equation, eln(x21)=e3.

Use the law of logarithm, elnx=x,x>0 and simplify further as follows.

x21=e3x2=e3+1x=±e3+1

Thus, the values of x are x=±e3+1.

(b)

To determine

### To solve: The equation, e2x−3ex+2=0 for x.

Expert Solution

The solutions for the given equation are x=0 and ln 2.

### Explanation of Solution

The given equation is e2x3ex+2=0.

Rewrite the given equation as (ex)23ex+2=0.

Substitute, ex=t and obtain the quadratic equation, t23t+2=0.

Find the factors of the quadratic equation.

t23t+2=0t2t2t+2=0(t1)(t2)=0

Thus, the value of t are t=1,2.

Substitute the values of t in t=ex, ex=1,ex=2.

Note that ex=b can be expressed as, b=lnb.

Therefore, x=ln1 and x=ln2.

Since ln 1 = 0, the values of x=0 and ln 2.

Therefore, the solutions for the given equation are x=0 and ln 2.

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