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

solve log_((24sinx)) (24cosx)=(3/2)

$$\:\:{solve} \\ $$$$\boldsymbol{{log}}_{\left(\mathrm{24}\boldsymbol{{sinx}}\right)} \left(\mathrm{24}\boldsymbol{{cosx}}\right)=\frac{\mathrm{3}}{\mathrm{2}} \\ $$

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

((ln24cosx)/(ln24sinx))=(3/2) ((ln24cosx)/3)=((ln24sinx)/2)=k 24cosx=e^(3k) 24sinx=e^(2k) 24^2 (cos^2 x+sin^2 x)=e^(6k) +e^(4k) (e^(2k) )^3 +(e^(2k) )^2 =576=8^3 +8^2 e^(2k) =8→2k=ln8 k=((3ln2)/2)=(3/2)ln2 now 24sinx=e^(2k) =8 sinx=(1/3) x=sin^(−1) ((1/3)) ★recheck...e^(2k) =8 e^k =(√8) e^(3k) =8(√8) so 24cosx=8(√8) cosx=((√8)/3)★

$$\frac{{ln}\mathrm{24}{cosx}}{{ln}\mathrm{24}{sinx}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\frac{{ln}\mathrm{24}{cosx}}{\mathrm{3}}=\frac{{ln}\mathrm{24}{sinx}}{\mathrm{2}}={k} \\ $$$$\mathrm{24}{cosx}={e}^{\mathrm{3}{k}} \\ $$$$\mathrm{24}{sinx}={e}^{\mathrm{2}{k}} \\ $$$$\mathrm{24}^{\mathrm{2}} \left({cos}^{\mathrm{2}} {x}+{sin}^{\mathrm{2}} {x}\right)={e}^{\mathrm{6}{k}} +{e}^{\mathrm{4}{k}} \\ $$$$\left({e}^{\mathrm{2}{k}} \right)^{\mathrm{3}} +\left({e}^{\mathrm{2}{k}} \right)^{\mathrm{2}} =\mathrm{576}=\mathrm{8}^{\mathrm{3}} +\mathrm{8}^{\mathrm{2}} \\ $$$${e}^{\mathrm{2}{k}} =\mathrm{8}\rightarrow\mathrm{2}\boldsymbol{{k}}=\boldsymbol{{ln}}\mathrm{8} \\ $$$$\boldsymbol{{k}}=\frac{\mathrm{3}\boldsymbol{{ln}}\mathrm{2}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}}\boldsymbol{{ln}}\mathrm{2} \\ $$$$\boldsymbol{{now}}\:\mathrm{24}\boldsymbol{{sinx}}=\boldsymbol{{e}}^{\mathrm{2}\boldsymbol{{k}}} =\mathrm{8} \\ $$$$\boldsymbol{{sinx}}=\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\:\:\boldsymbol{{x}}=\boldsymbol{{sin}}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$$ \\ $$$$ \\ $$$$\bigstar{recheck}...{e}^{\mathrm{2}{k}} =\mathrm{8}\:\:\:{e}^{{k}} =\sqrt{\mathrm{8}}\: \\ $$$${e}^{\mathrm{3}{k}} =\mathrm{8}\sqrt{\mathrm{8}}\:\:{so} \\ $$$$\mathrm{24}{cosx}=\mathrm{8}\sqrt{\mathrm{8}}\: \\ $$$${cosx}=\frac{\sqrt{\mathrm{8}}}{\mathrm{3}}\bigstar \\ $$

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

((ln (24 cos x))/(ln (24 sin x)))=(3/2) (24 cos x)^2 =(24 sin x)^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 ⇒s=sin x=(1/3) ⇒x=nπ+(−1)^n sin^(−1) (1/3) ⇒s=sin x=((−3±i(√(23)))/(16)) ⇒non−real

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

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

thk you sir

$${thk}\:{you}\:{sir} \\ $$

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

good very good

$${good}\:{very}\:{good} \\ $$

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

(24sin x)^(3/2) =24cos x (((sin x)^(3/2) )/(cos x))=((24)/(24^(3/2) )) (((sin x)^(3/2) )/(cos x))=((√6)/(12)) (((1−cos^2 x)^((1/2)×(3/2)) )/(cos x))=((√6)/(12)) ((((1−cos^2 x)^((1/2)×(3/2)) )^4 )/(cos^4 x))=((((√6))^4 )/(12^4 )) (((1−cos^2 x)^3 )/(cos^4 x))=(1/(576)) (((1−2cos^2 x+cos^4 x)(1−cos^2 x))/(cos^4 x))=(1/(576)) ((1−cos^2 x−2cos^2 x+2cos^4 x+cos^4 x−cos^6 x)/(cos^4 x))=(1/(576)) (1/(576)) 576−1728cos^2 x+1728cos^4 x−576cos^6 x− −cos^4 x=0 576cos^6 x−1727cos^4 x+1728cos^2 x−576=0 cos x=0.942809041582 x=0.339836 cos x=−0.942809041582 no more real solutions

$$\left(\mathrm{24sin}\:{x}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} =\mathrm{24cos}\:{x} \\ $$$$\frac{\left(\mathrm{sin}\:{x}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{cos}\:{x}}=\frac{\mathrm{24}}{\mathrm{24}^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$$\frac{\left(\mathrm{sin}\:{x}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{cos}\:{x}}=\frac{\sqrt{\mathrm{6}}}{\mathrm{12}} \\ $$$$\frac{\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {x}\right)^{\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{cos}\:{x}}=\frac{\sqrt{\mathrm{6}}}{\mathrm{12}} \\ $$$$\frac{\left(\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {x}\right)^{\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{2}}} \right)^{\mathrm{4}} }{\mathrm{cos}\:^{\mathrm{4}} {x}}=\frac{\left(\sqrt{\mathrm{6}}\right)^{\mathrm{4}} }{\mathrm{12}^{\mathrm{4}} } \\ $$$$\frac{\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {x}\right)^{\mathrm{3}} }{\mathrm{cos}\:^{\mathrm{4}} {x}}=\frac{\mathrm{1}}{\mathrm{576}} \\ $$$$\frac{\left(\mathrm{1}−\mathrm{2cos}\:^{\mathrm{2}} {x}+\mathrm{cos}\:^{\mathrm{4}} {x}\right)\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {x}\right)}{\mathrm{cos}\:^{\mathrm{4}} {x}}=\frac{\mathrm{1}}{\mathrm{576}} \\ $$$$\frac{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {x}−\mathrm{2cos}\:^{\mathrm{2}} {x}+\mathrm{2cos}\:^{\mathrm{4}} {x}+\mathrm{cos}\:^{\mathrm{4}} {x}−\mathrm{cos}\:^{\mathrm{6}} {x}}{\mathrm{cos}\:^{\mathrm{4}} {x}}=\frac{\mathrm{1}}{\mathrm{576}} \\ $$$$\frac{\mathrm{1}}{\mathrm{576}} \\ $$$$\mathrm{576}−\mathrm{1728cos}\:^{\mathrm{2}} {x}+\mathrm{1728cos}\:^{\mathrm{4}} {x}−\mathrm{576cos}\:^{\mathrm{6}} {x}− \\ $$$$−\mathrm{cos}\:^{\mathrm{4}} {x}=\mathrm{0} \\ $$$$\mathrm{576cos}\:^{\mathrm{6}} {x}−\mathrm{1727cos}\:^{\mathrm{4}} {x}+\mathrm{1728cos}\:^{\mathrm{2}} {x}−\mathrm{576}=\mathrm{0} \\ $$$$\mathrm{cos}\:{x}=\mathrm{0}.\mathrm{942809041582}\:\:\:\:\:{x}=\mathrm{0}.\mathrm{339836} \\ $$$$\mathrm{cos}\:{x}=−\mathrm{0}.\mathrm{942809041582}\:\: \\ $$$${no}\:{more}\:{real}\:{solutions} \\ $$$$ \\ $$

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