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Chapter 3: Problem 41

Solve the exponential equation algebraically. Approximate the result to three decimal places. $$\frac{500}{100-e^{x / 2}}=20$$

Short Answer

Expert verified
The solution for \(x\) is 8.318 when rounded to three decimal places.

Step by step solution

01

Simplify Equation

First, isolate the exponential term by simplifying the given equation. This is accomplished by multiplying both sides of the equation by \(100 - e^{x/2}\), leading to \(20(100 - e^{x/2}) = 500\).
02

Continue Simplification

Continuing with the simplification, distribute the 20 on the left-hand side to get \(2000 - 20e^{x/2} = 500\). Then, subtract 2000 from both sides to isolate the exponential term on one side, which results in \(-20e^{x/2} = -1500\).
03

Isolate the Exponential Term

To isolate the exponential term, divide both sides of the equation by -20. The equation then becomes \(e^{x/2} = 75\).
04

Apply Natural Logarithm to Both Sides

To free the variable \(x\) from the exponent, apply the natural logarithm to both sides of the equation. Remember that the inverse of the exponential function is the natural logarithm, so when we apply the natural logarithm to \(e^{x/2}\), we get \(x/2\). So the new equation becomes \(x/2 = ln(75)\).
05

Solve for x

Finally, finding \(x\) requires multiplying both sides of the equation by 2, which gives \(x = 2 * ln(75)\). This is approximately 8.31777.
06

Round to Three Decimal Places

The problem asks for the solution to be approximated to three decimal places, so \(x\) would be rounded to 8.318.

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Most popular questions from this chapter

A classmate claims that the following are true. (a) \(\ln (u+v)=\ln u+\ln v=\ln (u v)\) (b) \(\ln (u-v)=\ln u-\ln v=\ln \frac{u}{v}\) (c) \((\ln u)^{n}=n(\ln u)=\ln u^{n}\) Discuss how you would demonstrate that these claims are not true.

Use a graphing utility to graph and solve the equation. Approximate the result to three decimal places. Verify your result algebraically. $$\ln (x+1)=2-\ln x$$

Solve the logarithmic equation algebraically. Approximate the result to three decimal places. $$6 \log _{3}(0.5 x)=11$$

Determine whether the statement is true or false. Justify your answer. The domain of a logistic growth function cannot be the set of real numbers.

Use a graphing utility to graph the functions \(y_{1}=\ln x-\ln (x-3)\) and \(y_{2}=\ln \frac{x}{x-3}\) in the same viewing window. Does the graphing utility show the functions with the same domain? If so, should it? Explain your reasoning.
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