Discover a wealth of knowledge at Westonci.ca, where experts provide answers to your most pressing questions. Get quick and reliable solutions to your questions from a community of experienced professionals on our platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
To solve the equation
[tex]\[ \tan^{-1}(\ln(x^{2/3})) + \tan^{-1}(\ln(x^2)) + \tan^{-1}(\ln(x^2)) = \pi/2, \][/tex]
we start by simplifying each logarithmic term inside the inverse tangent functions.
Recall the logarithm properties: [tex]\( \ln(a^b) = b \ln(a) \)[/tex]. Applying this property:
1. [tex]\( \ln(x^{2/3}) = \frac{2}{3} \ln(x) \)[/tex]
2. [tex]\( \ln(x^2) = 2 \ln(x) \)[/tex]
Now rewrite the original equation using these logarithmic simplifications:
[tex]\[ \tan^{-1}\left(\frac{2}{3} \ln(x)\right) + \tan^{-1}(2 \ln(x)) + \tan^{-1}(2 \ln(x)) = \pi/2 \][/tex]
By combining the two identical terms:
[tex]\[ \tan^{-1}\left(\frac{2}{3} \ln(x)\right) + 2 \tan^{-1}(2 \ln(x)) = \pi/2 \][/tex]
Next, let [tex]\( y = \ln(x) \)[/tex]. Substituting [tex]\( y \)[/tex] into the equation:
[tex]\[ \tan^{-1}\left(\frac{2}{3} y\right) + 2 \tan^{-1}(2y) = \pi/2 \][/tex]
To simplify, assume [tex]\( a = \tan^{-1}\left(\frac{2}{3} y\right) \)[/tex] and [tex]\( b = \tan^{-1}(2y) \)[/tex]. Hence, the equation can be written as:
[tex]\[ a + 2b = \pi/2 \][/tex]
We use the identity for the sum of tangents for [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
[tex]\[ \tan(a + b) = \frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)} \][/tex]
Since [tex]\( \tan^{-1}(u) = a \)[/tex] implies [tex]\( \tan(a) = u \)[/tex]:
[tex]\[ \tan(a) = \frac{2}{3}y \quad \text{and} \quad \tan(b) = 2y \][/tex]
First, solve for [tex]\( \tan(a + b) \)[/tex]:
[tex]\[ \tan\left(a + 2b\right) = \tan\left(\frac{\pi}{2}\right) \rightarrow \text{undefined (since } \tan(\frac{\pi}{2}) \text{ is undefined)} \][/tex]
Therefore, for a valid solution as [tex]\( y \)[/tex]:
[tex]\[ a + 2b = \pi/2 \][/tex]
Let's assume simplifications to find [tex]\( y \)[/tex]:
Set [tex]\( \tan(a + b) = 1 \)[/tex]:
[tex]\[ \tan\left(\frac{\pi}{4}\right) = \frac{2y/3 + 2y}{1 - (2y/3)(2y)} = 1 \][/tex]
[tex]\[ 1 = \frac{(2y/3) + 2y}{1 - \frac{4y^2}{3}} \][/tex]
Solving this rational equation for [tex]\( y \)[/tex]:
[tex]\[ 1 = \frac{2y/3 + 2y}{1 - \frac{8y^2}{3}} \][/tex]
Cross-multiplying:
[tex]\[ 1 - \frac{8y^2}{3} = \frac{2y/3 + 2y} \][/tex]
Simplify and solve for [tex]\( y \)[/tex]:
[tex]\[ 1 - \frac{8y^2}{3} = \frac{8y/3} \][/tex]
Setting everything on one side:
[tex]\[ 1 = \frac{8y}{3} + \frac{8y^2}{3} \][/tex]
Multiply by 3:
[tex]\[ 3 = 8y + 8y^2 \][/tex]
Rearrange:
[tex]\[ 8y^2 + 8y - 3 = 0 \][/tex]
Solve this quadratic equation using the quadratic formula [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] where [tex]\( a = 8 \)[/tex], [tex]\( b = 8 \)[/tex], and [tex]\( c = -3 \)[/tex]:
[tex]\[ y = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 8 \cdot (-3)}}{2 \cdot 8} \][/tex]
[tex]\[ y = \frac{-8 \pm \sqrt{64 + 96}}{16} \][/tex]
[tex]\[ y = \frac{-8 \pm \sqrt{160}}{16} \][/tex]
Simplify:
[tex]\[ y = \frac{-8 \pm 4\sqrt{10}}{16} \][/tex]
[tex]\[ y = \frac{-1 \pm \sqrt{10}}{4} \][/tex]
Only consider positive solutions for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{\sqrt{10} - 1}{4} \][/tex]
Recall [tex]\( y = \ln(x) \)[/tex]:
[tex]\[ \ln(x) = \frac{\sqrt{10} - 1}{4} \][/tex]
Exponentiate both sides to solve for [tex]\( x \)[/tex]:
[tex]\[ x = e^{\frac{\sqrt{10} - 1}{4}} \][/tex]
Thus, the solution is:
[tex]\[ \boxed{e^{\frac{\sqrt{10} - 1}{4}}} \][/tex]
[tex]\[ \tan^{-1}(\ln(x^{2/3})) + \tan^{-1}(\ln(x^2)) + \tan^{-1}(\ln(x^2)) = \pi/2, \][/tex]
we start by simplifying each logarithmic term inside the inverse tangent functions.
Recall the logarithm properties: [tex]\( \ln(a^b) = b \ln(a) \)[/tex]. Applying this property:
1. [tex]\( \ln(x^{2/3}) = \frac{2}{3} \ln(x) \)[/tex]
2. [tex]\( \ln(x^2) = 2 \ln(x) \)[/tex]
Now rewrite the original equation using these logarithmic simplifications:
[tex]\[ \tan^{-1}\left(\frac{2}{3} \ln(x)\right) + \tan^{-1}(2 \ln(x)) + \tan^{-1}(2 \ln(x)) = \pi/2 \][/tex]
By combining the two identical terms:
[tex]\[ \tan^{-1}\left(\frac{2}{3} \ln(x)\right) + 2 \tan^{-1}(2 \ln(x)) = \pi/2 \][/tex]
Next, let [tex]\( y = \ln(x) \)[/tex]. Substituting [tex]\( y \)[/tex] into the equation:
[tex]\[ \tan^{-1}\left(\frac{2}{3} y\right) + 2 \tan^{-1}(2y) = \pi/2 \][/tex]
To simplify, assume [tex]\( a = \tan^{-1}\left(\frac{2}{3} y\right) \)[/tex] and [tex]\( b = \tan^{-1}(2y) \)[/tex]. Hence, the equation can be written as:
[tex]\[ a + 2b = \pi/2 \][/tex]
We use the identity for the sum of tangents for [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
[tex]\[ \tan(a + b) = \frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)} \][/tex]
Since [tex]\( \tan^{-1}(u) = a \)[/tex] implies [tex]\( \tan(a) = u \)[/tex]:
[tex]\[ \tan(a) = \frac{2}{3}y \quad \text{and} \quad \tan(b) = 2y \][/tex]
First, solve for [tex]\( \tan(a + b) \)[/tex]:
[tex]\[ \tan\left(a + 2b\right) = \tan\left(\frac{\pi}{2}\right) \rightarrow \text{undefined (since } \tan(\frac{\pi}{2}) \text{ is undefined)} \][/tex]
Therefore, for a valid solution as [tex]\( y \)[/tex]:
[tex]\[ a + 2b = \pi/2 \][/tex]
Let's assume simplifications to find [tex]\( y \)[/tex]:
Set [tex]\( \tan(a + b) = 1 \)[/tex]:
[tex]\[ \tan\left(\frac{\pi}{4}\right) = \frac{2y/3 + 2y}{1 - (2y/3)(2y)} = 1 \][/tex]
[tex]\[ 1 = \frac{(2y/3) + 2y}{1 - \frac{4y^2}{3}} \][/tex]
Solving this rational equation for [tex]\( y \)[/tex]:
[tex]\[ 1 = \frac{2y/3 + 2y}{1 - \frac{8y^2}{3}} \][/tex]
Cross-multiplying:
[tex]\[ 1 - \frac{8y^2}{3} = \frac{2y/3 + 2y} \][/tex]
Simplify and solve for [tex]\( y \)[/tex]:
[tex]\[ 1 - \frac{8y^2}{3} = \frac{8y/3} \][/tex]
Setting everything on one side:
[tex]\[ 1 = \frac{8y}{3} + \frac{8y^2}{3} \][/tex]
Multiply by 3:
[tex]\[ 3 = 8y + 8y^2 \][/tex]
Rearrange:
[tex]\[ 8y^2 + 8y - 3 = 0 \][/tex]
Solve this quadratic equation using the quadratic formula [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] where [tex]\( a = 8 \)[/tex], [tex]\( b = 8 \)[/tex], and [tex]\( c = -3 \)[/tex]:
[tex]\[ y = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 8 \cdot (-3)}}{2 \cdot 8} \][/tex]
[tex]\[ y = \frac{-8 \pm \sqrt{64 + 96}}{16} \][/tex]
[tex]\[ y = \frac{-8 \pm \sqrt{160}}{16} \][/tex]
Simplify:
[tex]\[ y = \frac{-8 \pm 4\sqrt{10}}{16} \][/tex]
[tex]\[ y = \frac{-1 \pm \sqrt{10}}{4} \][/tex]
Only consider positive solutions for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{\sqrt{10} - 1}{4} \][/tex]
Recall [tex]\( y = \ln(x) \)[/tex]:
[tex]\[ \ln(x) = \frac{\sqrt{10} - 1}{4} \][/tex]
Exponentiate both sides to solve for [tex]\( x \)[/tex]:
[tex]\[ x = e^{\frac{\sqrt{10} - 1}{4}} \][/tex]
Thus, the solution is:
[tex]\[ \boxed{e^{\frac{\sqrt{10} - 1}{4}}} \][/tex]
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.
