find X
- Thread starter mathe25
- Start date
Hello, mathe25!
I hope I read it correctly . . . It is a very unpleasant problem.
I reached a point where I don't want to go any further . . . any takers?
\(\displaystyle \text{We have: }\:9\,\dfrac{\ln(\cos^2\!x)}{\ln(\sin2x)} + 8\,\dfrac{\ln(\sin x)}{\ln(2\cos x)} \:=\:16 \)
Multiply by \(\displaystyle \ln(\sin2x)\ln(2\cos x):\)
\(\displaystyle 9\ln(\cos^2\!x)\ln(2\cos x) + 8\ln(\sin x)\ln(\sin2x) \;=\;16\ln(\sin2x)\ln(2\cos x) \)
\(\displaystyle 9\ln(\cos^2x)\ln(2\cos x) + 8\ln(\sin x)\ln(2\sin x\cos x) \;=\) .\(\displaystyle 16\ln(2\sin x\cos x)\ln(2\cos x) \)
\(\displaystyle 9\cdot2\ln(\cos x)\big[\ln(2) + \ln(\cos x)\big] + 8\ln(\sin x)\big[\ln(2) + \ln(\sin x) + \ln(\cos x)\big]\)
. . . \(\displaystyle =\; 16\big[\ln(2) + \ln(\sin x) + \ln(\cos x)\big]\,\big[\ln(2) + \ln(\cos x)\big] \)
\(\displaystyle 18\ln(2)\ln(\cos x) + 18\big[\ln(\cos x)\big]^2 + 8\ln(2)\ln(\sin x) +\) \(\displaystyle 8\big[\ln{\sin x)\big]^2 + 8\ln(\sin x)\ln(\cos x)\)
. . . \(\displaystyle =\;16\big[\ln(2)\big]^2 + 16\ln(2)\ln(\cos x) + 16\ln(2)\ln(\sin x) + 16\ln(\sin x)\ln(\cos x) \)
. . . . . . \(\displaystyle + 16\ln(2)\ln(\cos x) + 16\big[\ln(\cos x)\big]^2\)
\(\displaystyle 2\big[\ln(\cos x)\big]^2\!-\!8\ln(\sin x)\ln(\cos x)\!+\!8\big[\ln(\sin x)\big]^2\!-\!14\ln(2)\ln(\cos x) -\) \(\displaystyle 8\ln(2)\ln(\sin x)\!-\!16\big[\ln(2)\big]^2 \,=\:0 \)
This is where I fell asleep . . .
I hope I read it correctly . . . It is a very unpleasant problem.
I reached a point where I don't want to go any further . . . any takers?
\(\displaystyle \text{We have: }\:9\,\dfrac{\ln(\cos^2\!x)}{\ln(\sin2x)} + 8\,\dfrac{\ln(\sin x)}{\ln(2\cos x)} \:=\:16 \)
Multiply by \(\displaystyle \ln(\sin2x)\ln(2\cos x):\)
\(\displaystyle 9\ln(\cos^2\!x)\ln(2\cos x) + 8\ln(\sin x)\ln(\sin2x) \;=\;16\ln(\sin2x)\ln(2\cos x) \)
\(\displaystyle 9\ln(\cos^2x)\ln(2\cos x) + 8\ln(\sin x)\ln(2\sin x\cos x) \;=\) .\(\displaystyle 16\ln(2\sin x\cos x)\ln(2\cos x) \)
\(\displaystyle 9\cdot2\ln(\cos x)\big[\ln(2) + \ln(\cos x)\big] + 8\ln(\sin x)\big[\ln(2) + \ln(\sin x) + \ln(\cos x)\big]\)
. . . \(\displaystyle =\; 16\big[\ln(2) + \ln(\sin x) + \ln(\cos x)\big]\,\big[\ln(2) + \ln(\cos x)\big] \)
\(\displaystyle 18\ln(2)\ln(\cos x) + 18\big[\ln(\cos x)\big]^2 + 8\ln(2)\ln(\sin x) +\) \(\displaystyle 8\big[\ln{\sin x)\big]^2 + 8\ln(\sin x)\ln(\cos x)\)
. . . \(\displaystyle =\;16\big[\ln(2)\big]^2 + 16\ln(2)\ln(\cos x) + 16\ln(2)\ln(\sin x) + 16\ln(\sin x)\ln(\cos x) \)
. . . . . . \(\displaystyle + 16\ln(2)\ln(\cos x) + 16\big[\ln(\cos x)\big]^2\)
\(\displaystyle 2\big[\ln(\cos x)\big]^2\!-\!8\ln(\sin x)\ln(\cos x)\!+\!8\big[\ln(\sin x)\big]^2\!-\!14\ln(2)\ln(\cos x) -\) \(\displaystyle 8\ln(2)\ln(\sin x)\!-\!16\big[\ln(2)\big]^2 \,=\:0 \)
This is where I fell asleep . . .
Hello, mathe25!
Okay, that helps a bit . . . but it's still an ugly mess.
This was the 13th Labor of Hercules . . .
We have: .\(\displaystyle 9\log_{\sin2x}(2\cos x)^2 + 8\log_{2\cos x}(\sin x) \;=\;16\)
. . . . \(\displaystyle 18\log_{\sin2x}(2\cos x) + 8\log_{2\cos x}(\sin x) \;=\;16\)
. . . . . . \(\displaystyle 18\,\dfrac{\log(2\cos x)}{\log(\sin2x)} + 8\,\dfrac{\log(\sin x)}{\log(2\cos x)} \;=\;16\)
Multiply by \(\displaystyle \log(\sin2x)\log(2\cos x)\!:\)
. . \(\displaystyle 18\big[\log(2\cos x)\big]^2 + 8\log(\sin x)\log(\sin2x) \;=\;16\log(\sin2x)\log(2\cos x)\)
. . \(\displaystyle 18\big[\log(2\cos x)\big]^2 + 8\log(\sin x)\log(2\sin x\cos x)\) .\(\displaystyle =\;16\log(2\sin x\cos x)\log(2\cos x)\)
. . \(\displaystyle 18\big[\log(2) + \log(\cos x)\big]^2 + 8\log(\sin x)\big[\log(2) + \log(\sin x) + \log(\cos x)\big]\)
. . . . . . \(\displaystyle = \;16\big[\log(2) + \log(\sin x) + \log(\cos x)\big]\,\big[\log(2) + \log(\cos x)\big] \)
\(\displaystyle \text{Let: }\:\begin{Bmatrix} T &=& \log(2) \\ C &=& \log(\cos x) \\ S &=& \log(\sin x}\end{Bmatrix}\)
We have: .\(\displaystyle 18(T+C)^2 + 8S(T+S+C) \;=\;16(T+S+C)(T+C)\)
. . \(\displaystyle 18T^2 + 36TC + 18C^2 + 8TS + 8S^2 + 8CS\)
. . . . . . \(\displaystyle =\;16T^2 + 16TC + 16TS + 16CS + 16TC + 16C^2\)
. . \(\displaystyle 2C^2 - 8CS + 8S^2 + 4TC - 8TS + 2T^2 \;=\;0\)
. . .\(\displaystyle C^2 - 4CS + 4S^2 + 2TC - 4TS + T^2 \;=\;0\)
. . . . \(\displaystyle (C-2S)^2 + 2T(C-2S) + T^2 \;=\;0\)
. . . . . . . . . \(\displaystyle \big[(C-2S) + T\big]^2 \;=\;0\)
. . . . . . . . . . .\(\displaystyle C - 2S + T \;=\;0\)
Back-substitute: .\(\displaystyle \log(\cos x) - 2\log(\sin x) + \log(2) \;=\;0\)
. . . . . . . . . . . . . . \(\displaystyle \log(\cos x) - \log(\sin^2\!x) \,=\,\text{-}\log(2)\)
. . . . . . . . . . . . . . . . . \(\displaystyle \log\left(\dfrac{\cos x}{\sin^2\!x}\right) \;=\;\log\left(2^{\text{-}1}\right)\)
And we have: .\(\displaystyle \dfrac{\cos x}{\sin^2\!x} \;=\;\dfrac{1}{2} \quad\Rightarrow\quad 2\cos x \;=\;\sin^2\!x \)
. . \(\displaystyle 2\cos x \;=\;1-\cos^2\!x \quad\Rightarrow\quad \cos^2\!x + 2\cos x - 1 \;=\;0\)
Quadratic Formula: .\(\displaystyle \cos x \;=\;\dfrac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2(1)} \;=\;-1 \pm\sqrt{2}\)
Since \(\displaystyle \cos x\) must be positive: .\(\displaystyle \cos x \:=\:\sqrt{2}-1\)
Therefore: .\(\displaystyle x \;=\;\cos^{\text{-}1}\!\left(\sqrt{2}-1\right) \;=\;1.1437177\hdots \)
. . . *pant* ... *gasp* ... Whew!
Okay, that helps a bit . . . but it's still an ugly mess.
This was the 13th Labor of Hercules . . .
We have: .\(\displaystyle 9\log_{\sin2x}(2\cos x)^2 + 8\log_{2\cos x}(\sin x) \;=\;16\)
. . . . \(\displaystyle 18\log_{\sin2x}(2\cos x) + 8\log_{2\cos x}(\sin x) \;=\;16\)
. . . . . . \(\displaystyle 18\,\dfrac{\log(2\cos x)}{\log(\sin2x)} + 8\,\dfrac{\log(\sin x)}{\log(2\cos x)} \;=\;16\)
Multiply by \(\displaystyle \log(\sin2x)\log(2\cos x)\!:\)
. . \(\displaystyle 18\big[\log(2\cos x)\big]^2 + 8\log(\sin x)\log(\sin2x) \;=\;16\log(\sin2x)\log(2\cos x)\)
. . \(\displaystyle 18\big[\log(2\cos x)\big]^2 + 8\log(\sin x)\log(2\sin x\cos x)\) .\(\displaystyle =\;16\log(2\sin x\cos x)\log(2\cos x)\)
. . \(\displaystyle 18\big[\log(2) + \log(\cos x)\big]^2 + 8\log(\sin x)\big[\log(2) + \log(\sin x) + \log(\cos x)\big]\)
. . . . . . \(\displaystyle = \;16\big[\log(2) + \log(\sin x) + \log(\cos x)\big]\,\big[\log(2) + \log(\cos x)\big] \)
\(\displaystyle \text{Let: }\:\begin{Bmatrix} T &=& \log(2) \\ C &=& \log(\cos x) \\ S &=& \log(\sin x}\end{Bmatrix}\)
We have: .\(\displaystyle 18(T+C)^2 + 8S(T+S+C) \;=\;16(T+S+C)(T+C)\)
. . \(\displaystyle 18T^2 + 36TC + 18C^2 + 8TS + 8S^2 + 8CS\)
. . . . . . \(\displaystyle =\;16T^2 + 16TC + 16TS + 16CS + 16TC + 16C^2\)
. . \(\displaystyle 2C^2 - 8CS + 8S^2 + 4TC - 8TS + 2T^2 \;=\;0\)
. . .\(\displaystyle C^2 - 4CS + 4S^2 + 2TC - 4TS + T^2 \;=\;0\)
. . . . \(\displaystyle (C-2S)^2 + 2T(C-2S) + T^2 \;=\;0\)
. . . . . . . . . \(\displaystyle \big[(C-2S) + T\big]^2 \;=\;0\)
. . . . . . . . . . .\(\displaystyle C - 2S + T \;=\;0\)
Back-substitute: .\(\displaystyle \log(\cos x) - 2\log(\sin x) + \log(2) \;=\;0\)
. . . . . . . . . . . . . . \(\displaystyle \log(\cos x) - \log(\sin^2\!x) \,=\,\text{-}\log(2)\)
. . . . . . . . . . . . . . . . . \(\displaystyle \log\left(\dfrac{\cos x}{\sin^2\!x}\right) \;=\;\log\left(2^{\text{-}1}\right)\)
And we have: .\(\displaystyle \dfrac{\cos x}{\sin^2\!x} \;=\;\dfrac{1}{2} \quad\Rightarrow\quad 2\cos x \;=\;\sin^2\!x \)
. . \(\displaystyle 2\cos x \;=\;1-\cos^2\!x \quad\Rightarrow\quad \cos^2\!x + 2\cos x - 1 \;=\;0\)
Quadratic Formula: .\(\displaystyle \cos x \;=\;\dfrac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2(1)} \;=\;-1 \pm\sqrt{2}\)
Since \(\displaystyle \cos x\) must be positive: .\(\displaystyle \cos x \:=\:\sqrt{2}-1\)
Therefore: .\(\displaystyle x \;=\;\cos^{\text{-}1}\!\left(\sqrt{2}-1\right) \;=\;1.1437177\hdots \)
. . . *pant* ... *gasp* ... Whew!
D
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.I looked for x, but it's too dark: couldn't find it....... I found it!!!!
................↑.........................right here