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Chapter 4: Problem 10

Find a number \(x\) such that \(\ln x=-3\).

Short Answer

Expert verified
The value of \(x\) that satisfies the equation \(\ln x = -3\) is approximately \(0.0498\).

Step by step solution

01

Rewrite the natural logarithm equation as an exponential equation

Recall that the natural logarithm function is the inverse function of the exponential function, \(e^x\). Therefore, we can rewrite the equation \(\ln x = -3\) as an exponential equation: \[e^{-3} = x\]
02

Evaluate the exponential function

Now that we have an exponential equation, we can evaluate the exponential function: \[x = e^{-3}\]
03

Calculate the value of x

Using a calculator to find the value of \(e^{-3}\), we get: \[x \approx 0.0498\] So, the value of \(x\) that satisfies the equation \(\ln x = -3\) is approximately \(0.0498\).

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Most popular questions from this chapter

Using a calculator, discover a formula for a good approximation for $$ \ln (2+t)-\ln 2 $$ for small values of \(t\) (for example, try \(t=0.04\), \(t=0.02, t=0.01,\) and then smaller values of \(t\) ). Then explain why your formula is indeed a good approximation.

Explain why the two previous problems imply that \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\) is the midpoint of the line segment with endpoints \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\)

Suppose one bank account pays \(3 \%\) annual interest compounded once per year, and a second bank account pays \(4 \%\) annual interest compounded continuously. If both bank accounts start with the same initial amount, how long will it take for the second bank account to contain \(50 \%\) more than the first bank account?

Suppose a colony of bacteria has tripled in two hours. What is the continuous growth rate of this colony of bacteria?

About how many years does it take for \(\$ 200\) to become \(\$ 800\) when compounded continuously at \(2 \%\) per year?
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