# The equation, e x = 5 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, Problem 26RE

(a)

To determine

## To solve: The equation, ex=5 for x.

Expert Solution

Solution:

The value of x=ln5.

### Explanation of Solution

The given equation is ex=5.

Take the natural logarithm on both the sides, ln(ex)=ln5.

Use the law of logarithm, ln(ex)=x,x express the above equation as x=ln5.

Therefore, the value of x=ln5.

(b)

To determine

### To solve: The equation, lnx=2 for x.

Expert Solution

Solution:

The value of x=e2.

### Explanation of Solution

The given equation is, lnx=2.

Take the natural logarithm on both the sides, elnx=e2.

Use the law of logarithm elnx=x,x>0, it can be simplified as, x=e2.

Thus, the value of x=e2.

(c)

To determine

### To solve: The equation, eex=2 for x.

Expert Solution

Solution:

The value of x=ln(ln2).

### Explanation of Solution

The given equation is eex=2.

Take the natural logarithm on both the sides, ln(eex)=ln2

Use the law of logarithm, ln(ex)=x,x twice and simplify as follows,

ln(eex)=ln2ex=ln2ln(ex)=ln(ln2)x=ln(ln2).

Therefore, the value of x=ln(ln2).

(d)

To determine

### To solve: The equation, tan−1x=1 for x.

Expert Solution

Solution:

The value of x=tan1.

### Explanation of Solution

The given equation is, tan1x=1.

Express the given equation as, x=tan1

Therefore, the value of x=tan1.

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