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Chapter 4: Problem 45

Solve each exponential equation. Express the solution set in terms of natural logarithms or common logarithms. Then use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. $$e^{4 x}+5 e^{2 x}-24=0$$

Short Answer

Expert verified
The only solution is \( x \approx 0.55 \)

Step by step solution

01

Express as Quadratic Equation

Let \(y = e^{2x}\). Hence the equation \(e^{4 x}+5 e^{2 x}-24=0\) becomes \(y^2 + 5y - 24 = 0\)
02

Use Quadratic Formula

Now, use the quadratic formula to solve for y, where \(a = 1\), \(b = 5\), \(c = -24\). Therefore, the solution for \(y\) will be \[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{25 - -96}}{2} = \frac{-5 \pm \sqrt{121}}{2} = \frac{-5 \pm 11}{2}\]
03

Solve for y

Hence, the two solutions for \(y\) are \[y = 3, y = -8\]
04

Back Substitution

Substitute \(y = e^{2x}\) back into the result: \(e^{2x} = 3\) and \(e^{2x} = -8\).
05

Applying Natural Logarithms

We know that \(e^{2x} = -8\) has no solution because the exponential function is always positive. For the other solution: Apply the natural logarithm to both sides, \(ln(e^{2x}) = ln(3)\), which simplifies to \(2x = ln(3)\). Divide by 2: \(x = \frac{ln(3)}{2}\)
06

Decimal Approximation

Finally, calculate a decimal approximation using a calculator: \( x \approx 0.55 \)

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Most popular questions from this chapter

What question can be asked to help evaluate \(\log _{3} 81 ?\)

Evaluate the indicated logarithmic expressions without using a calculator. a. Evaluate: \(\log _{2} 16\) b. Evaluate: \(\log _{2} 32-\log _{2} 2\) c. What can you conclude about $$\log _{2} 16, \text { or } \log _{2}\left(\frac{32}{2}\right) ?$$

Graph \(f\) and \(g\) in the same viewing rectangle. Then describe the relationship of the graph of g to the graph of \(f\). $$f(x)=\ln x, g(x)=\ln x+3$$

Determine whether each statement makes sense or does not make sense, and explain your reasoning. I've noticed that exponential functions and logarithmic functions exhibit inverse, or opposite, behavior in many ways. For example, a vertical translation shifts an exponential function's horizontal asymptote and a horizontal translation shifts a logarithmic function's vertical asymptote.

Determine whether each statement makes sense or does not make sense, and explain your reasoning. I estimate that \(\log _{8} 16\) lies between 1 and 2 because \(8^{1}=8\) and \(8^{2}=64\).
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