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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 5: Q 110. (page 299)

Solve the equation.
$4 e^{x + 1} = 5$

Short Answer

Expert verified

The solution to the equation, $4 e^{x + 1} = 5$ is, $x = \ln \frac{5}{4} - 1$ .

Step by step solution

01

Step 1. Given Information

The given equation to be solved is $4 e^{x + 1} = 5$ .

02

Step 2. Formula used

Natural logarithm:

$\ln e = 1$

03

Step 3. Finding Solution

The given equation is $4 e^{x + 1} = 5$ .

Divide both sides by $4$ .

$\frac{4 e^{x + 1}}{4} = \frac{5}{4} e^{x + 1} = \frac{5}{4}$

Take logarithm on both sides.

$\ln e^{x + 1} = \ln \frac{5}{4} (x + 1) \ln e = \ln \frac{5}{4} x + 1 = \ln \frac{5}{4} x = \ln \frac{5}{4} - 1$

The solution to the given equationis, $x = \ln \frac{5}{4} - 1$ .

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Most popular questions from this chapter

Evaluate each expression using the values given in the table.

(a) $(f 鈭� g) (1)$ (b) $(f 鈭� g) (2)$ (c) $(g 鈭� f) (2)$ (d) $(g 鈭� f) (3)$ (e) $(g 鈭� g) (1)$ (f) $(f 鈭� f) (3)$ .

The domain of one-to-one function g is $[0, 15]$ , and range is $(0, 8)$ . State the domain and range oflocalid="1646180141229" $g^{- 1}$

Use the graph of $y = f (x)$ given in the Problem $43$ to evaluate the following:

(a) $f (- 1)$

(b) $f (1)$

(c) $f^{- 1} (1)$

(d) $f^{- 1} (2)$

the function f is one-to-one. Find its inverse and check your answer.

$f (x) = \frac{x^{2} - 4}{2 x^{2}} x > 0$

Begin with the graph of $y = e^{x}$ and use transformation to graph the function. Determine the domain, range and horizontal asymptote of the function.

localid="1646160286433" $f (x) = 5 - e^{- x}$

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