Answer :
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]
