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Chapter 3: Problem 13

Find all numbers \(y\) such that \(\ln \left(y^{2}+1\right)=3\).

Short Answer

Expert verified
The two solutions for y are: \(y \approx 4.367\) and \(y \approx -4.367\).

Step by step solution

01

Write down the given equation

We are given the equation: \(\ln \left(y^{2} + 1\right) = 3\)
02

Exponentiate both sides of the equation

To get rid of the natural logarithm, we have to exponentiate both sides of the equation. Using the property of logarithms that states that \(a = \ln b\) if and only if \(e^a = b\), we get: \(e^{\ln \left(y^{2} + 1\right)} = e^{3}\)
03

Simplify the left-hand side

Now we can simplify the left-hand side of the equation using the property of logarithms that states that \(e^{\ln a} = a\). So, we have: \(y^{2} + 1 = e^3\)
04

Subtract 1 from both sides

To make the equation easier to solve, we'll subtract 1 from both sides: \(y^{2} + 1 - 1 = e^{3} - 1\). It simplifies to: \(y^{2} = e^{3} - 1\)
05

Take the square root of both sides

To solve for y, we need to take the square root of both sides of the equation. Remember that we need to consider both the positive and negative square roots: \(y = \pm\sqrt{e^3 - 1}\)
06

Evaluate

Finally, we can evaluate the positive and negative square roots: \(y = \pm\sqrt{e^3 - 1} = \pm\sqrt{20.0855369 - 1} = \pm\sqrt{19.0855369}\) The two solutions for y are: \(y \approx 4.367\) and \(y \approx -4.367\)

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