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Chapter 3: Problem 40

Solve the exponential equation algebraically. Approximate the result to three decimal places. $$e^{2 x}-5 e^{x}+6=0$$

Short Answer

Expert verified
The solutions to the equation are \(x = ln(2) \approx 0.693\) and \(x = ln(3) \approx 1.099\) (with approximation to 3 decimal places)

Step by step solution

01

Express the Equation in Quadratic Form

Let's replace \(e^{x}\) with \(y\). So the equation becomes \(y^2 - 5y + 6 = 0\), which is a quadratic equation and can be solved using factorization or quadratic formula.
02

Solve the Quadratic Equation

Using factorization method, the equation \(y^2 - 5y + 6 = 0\) can be written as \((y - 2)(y - 3) = 0\). Setting each factor equal to zero gives the solutions \(y_1 = 2\) and \(y_2 = 3\).
03

Substitute y Back Into the Equation

Now let's substitute \(y = e^{x}\) back into the solutions. So, our solutions become \(e^{x} = 2\) and \(e^{x} = 3\).
04

Solve for x

To solve for x, take the natural logarithm on both sides of the equations. This gives us \(x = ln(2)\) and \(x = ln(3)\).

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Most popular questions from this chapter

In Exercises \(97-102,\) determine whether the statement is true or false given that \(f(x)=\ln x .\) Justify your answer. $$\sqrt{f(x)}=\frac{1}{2} f(x)$$

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