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Chapter 8: Problem 2

Solve (a) \(\ln x=2.4050\) (b) \(\ln x=0.9611\) (c) \(\ln x=-0.9611\) (d) \(\ln x=-2.0000\)

Short Answer

Expert verified
Question: Find the value of x for each natural logarithm equation. (a) \(\ln x = 2.4050\) (b) \(\ln x = 0.9611\) (c) \(\ln x = -0.9611\) (d) \(\ln x = -2.0000\) Answer: (a) \(x \approx 11.077\) (b) \(x \approx 2.616\) (c) \(x \approx 0.382\) (d) \(x \approx 0.135\)

Step by step solution

01

Understand the given equations

In the exercise, we are given four equations, each containing a natural logarithm of the variable x with a specific value. Our goal is to find the value of x for each equation.
02

Convert the equation to exponential form

To solve for x, we will convert each natural logarithm equation into an exponential form. The general formula for this conversion is: $$\ln(x) = a \Rightarrow x = e^a$$ We will apply this formula to each equation in the exercise.
03

Solve (a) \(\ln x=2.4050\)

Using the formula to convert to exponential form and solve for x: $$x = e^{2.4050}$$ Now, compute the value of x: $$x \approx 11.077$$
04

Solve (b) \(\ln x=0.9611\)

Again, use the formula to convert to exponential form and solve for x: $$x = e^{0.9611}$$ Now, compute the value of x: $$x \approx 2.616$$
05

Solve (c) \(\ln x=-0.9611\)

Use the formula to convert to exponential form and solve for x: $$x = e^{-0.9611}$$ Now, compute the value of x: $$x \approx 0.382$$
06

Solve (d) \(\ln x=-2.0000\)

Finally, use the formula to convert to exponential form and solve for x: $$x = e^{-2.0000}$$ Now, compute the value of x: $$x \approx 0.135$$ The solutions for the given equations are as follows: (a) \(x \approx 11.077\) (b) \(x \approx 2.616\) (c) \(x \approx 0.382\) (d) \(x \approx 0.135\)

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Most popular questions from this chapter

Calculate the voltage gain in decibels of an amplifier where the input signal is \(0.15 \mathrm{~V}\) and the output signal is \(1.9 \mathrm{~V}\).

Write the following using logarithms: (a) \(10^{2}=100\) (b) \(0.001=10^{-3}\) (c) \(\mathrm{e}^{-1.3}=0.2725\) (d) \(\mathrm{e}^{1.5}=4.4817\)

The current in a circuit, \(i(t)\), is given by $$ i(t)=25 \mathrm{e}^{-0.2 t} \quad t \geq 0 $$ (a) State the current when \(t=0\). (b) Calculate the value of the current when \(t=2\) (c) Calculate the time when the value of the current is \(12.5\).

The current, \(I\), through a diode is given by $$ I=I_{\mathrm{s}}\left(\mathrm{e}^{40 \mathrm{~V}}-1\right) $$ where \(I_{\mathrm{s}}\) is the reverse saturation current and \(V\) is the voltage across the diode. If the current is 300 times greater than the reverse saturation current, calculate the voltage across the diode.

Expand the brackets of the following expressions: (a) \(\left(e^{x}+2\right)^{2}\) (b) \(\left(\mathrm{e}^{x}+1\right)\left(\mathrm{e}^{-x}-1\right)\) (c) \(\left(\mathrm{e}^{2 x}+\mathrm{e}^{x}\right)\left(\mathrm{e}^{-2 x}+\mathrm{e}^{-x}\right)\) (d) \(\left(1+\mathrm{e}^{2 x}+\mathrm{e}^{-2 x}\right)\left(1-\mathrm{e}^{x}\right)\)
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