91Ó°ÊÓ

First, let's convert the given equation from logarithmic form to exponential form. The equation is: \[\ln(2r^2 - 3) = -1\] Since \(\ln(x)\) is the natural logarithm, which has the base \(e\), the equation can be rewritten in exponential form as: \[e^{-1} = 2r^2 - 3\]

Now, we need to isolate \(r\) to find its value(s). The equation we have is: \[e^{-1} = 2r^2 - 3\] Let's first isolate the quadratic term, \(2r^2\), by moving the constant term to the other side: \[2r^2 = 3+e^{-1}\] Next, divide by \(2\) to solve for \(r^2\): \[r^2 = \frac{3+e^{-1}}{2}\] Finally, take the square root of both sides to solve for \(r\). Since we took the square root, we should get two possible values for \(r\): \[r = \pm\sqrt{\frac{3+e^{-1}}{2}}\]

To ensure that our solutions are valid, we should substitute the values of \(r\) back into the original equation and ensure that they satisfy the equation. Plug both positive and negative solutions for \(r\) into our original equation \(\ln(2r^2 - 3) = -1\): Positive solution: \(r = \sqrt{\frac{3+e^{-1}}{2}}\) \[\ln\left(2\left(\sqrt{\frac{3+e^{-1}}{2}}\right)^2 - 3\right) = \ln(2\cdot\frac{3+e^{-1}}{2} - 3) = \ln(e^{-1}) = -1\] Negative solution: \(r = -\sqrt{\frac{3+e^{-1}}{2}}\) \[\ln\left(2\left(-\sqrt{\frac{3+e^{-1}}{2}}\right)^2 - 3\right) = \ln(2\cdot\frac{3+e^{-1}}{2} - 3) = \ln(e^{-1}) = -1\] Since both the positive and negative solutions satisfy the given equation, we can conclude that the solutions are: \[r = \pm\sqrt{\frac{3+e^{-1}}{2}}\]