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Question Number 83289 by Cmr 237 last updated on 29/Feb/20

$$solve$$$${\mathit{l}\mathit{o}\mathit{g}}_{\left(24\mathit{s}\mathit{i}\mathit{n}\mathit{x}\right)}\left(24\mathit{c}\mathit{o}\mathit{s}\mathit{x}\right)=\frac{3}{2}$$

Answered by TANMAY PANACEA last updated on 29/Feb/20

$$\frac{ln24cosx}{ln24sinx}=\frac{3}{2}$$$$\frac{ln24cosx}{3}=\frac{ln24sinx}{2}=k$$$$24cosx=e^{3k}$$$$24sinx=e^{2k}$$$${24}^2({cos}^{2}x+{sin}^{2}x)=e^{6k}+e^{4k}$$$${\left(e^{2k}\right)}^3+{\left(e^{2k}\right)}^2=576=8^3+8^2$$$$e^{2k}=8\to 2\mathit{k}=\mathit{l}\mathit{n}8$$$$\mathit{k}=\frac{3\mathit{l}\mathit{n}2}{2}=\frac{3}{2}\mathit{l}\mathit{n}2$$$$\mathit{n}\mathit{o}\mathit{w}24\mathit{s}\mathit{i}\mathit{n}\mathit{x}={\mathit{e}}^{2\mathit{k}}=8$$$$\mathit{s}\mathit{i}\mathit{n}\mathit{x}=\frac{1}{3}\mathit{x}={\mathit{s}\mathit{i}\mathit{n}}^{-1}\left(\frac{1}{3}\right)$$$$$$$$$$$$★recheck...e^{2k}=8e^k=\sqrt{8}$$$$e^{3k}=8\sqrt{8}so$$$$24cosx=8\sqrt{8}$$$$cosx=\frac{\sqrt{8}}{3}★$$

Answered by mr W last updated on 29/Feb/20

$$\frac{\ln \left(24\cos x\right)}{\ln \left(24\sin x\right)}=\frac{3}{2}$$$${\left(24\cos x\right)}^2={\left(24\sin x\right)}^3$$$${\cos }^{2}x=24{\sin }^{3}x$$$$24{\sin }^{3}x+{\sin }^{2}x-1=0$$$$24s^3+s^2-1=0$$$$(3s-1)(8s^2+3s+1)=0$$$$\Rightarrow s=\sin x=\frac{1}{3}\Rightarrow x=n\pi +{(-1)}^n{\sin }^{-1}\frac{1}{3}$$$$\Rightarrow s=\sin x=\frac{-3\pm i\sqrt{23}}{16}\Rightarrow non-real$$

Commented by Cmr 237 last updated on 29/Feb/20

$$thkyousir$$

Commented by jacoque@gmail.com last updated on 29/Feb/20

$$goodverygood$$

Answered by jacoque@gmail.com last updated on 29/Feb/20

$${\left(24\sin x\right)}^{\frac{3}{2}}=24\cos x$$$$\frac{{\left(\sin x\right)}^{\frac{3}{2}}}{\cos x}=\frac{24}{{24}^{\frac{3}{2}}}$$$$\frac{{\left(\sin x\right)}^{\frac{3}{2}}}{\cos x}=\frac{\sqrt{6}}{12}$$$$\frac{{(1-\cos {}^{2}x)}^{\frac{1}{2}\times \frac{3}{2}}}{\cos x}=\frac{\sqrt{6}}{12}$$$$\frac{{\left({(1-\cos {}^{2}x)}^{\frac{1}{2}\times \frac{3}{2}}\right)}^4}{\cos {}^{4}x}=\frac{{\left(\sqrt{6}\right)}^4}{{12}^4}$$$$\frac{{(1-\cos {}^{2}x)}^3}{\cos {}^{4}x}=\frac{1}{576}$$$$\frac{(1-2\cos {}^{2}x+\cos {}^{4}x)(1-\cos {}^{2}x)}{\cos {}^{4}x}=\frac{1}{576}$$$$\frac{1-\cos {}^{2}x-2\cos {}^{2}x+2\cos {}^{4}x+\cos {}^{4}x-\cos {}^{6}x}{\cos {}^{4}x}=\frac{1}{576}$$$$\frac{1}{576}$$$$576-1728\cos {}^{2}x+1728\cos {}^{4}x-576\cos {}^{6}x-$$$$-\cos {}^{4}x=0$$$$576\cos {}^{6}x-1727\cos {}^{4}x+1728\cos {}^{2}x-576=0$$$$\cos x=0.942809041582x=0.339836$$$$\cos x=-0.942809041582$$$$nomorerealsolutions$$$$$$

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