Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 167648 by mathlove last updated on 22/Mar/22

log_(sinx) 2+log_(cosx) 2+log_(sin x) 2×log_(cos x) 2=0  x=?

$$\mathrm{log}_{{sinx}} \mathrm{2}+\mathrm{log}_{{cosx}} \mathrm{2}+{log}_{\mathrm{sin}\:{x}} \mathrm{2}×{log}_{\mathrm{cos}\:{x}} \mathrm{2}=\mathrm{0} \\ $$$${x}=? \\ $$

Commented by Rasheed.Sindhi last updated on 22/Mar/22

log_(sinx) 2=y (say)  (sinx)^y =2  ∵ 0≤sinx≤1  ∴ (sinx)^y ≤1 ∀ y∈R  ∴ log_(sinx) 2 ∉R

$$\mathrm{log}_{{sinx}} \mathrm{2}={y}\:\left({say}\right) \\ $$$$\left({sinx}\right)^{{y}} =\mathrm{2} \\ $$$$\because\:\mathrm{0}\leqslant{sinx}\leqslant\mathrm{1} \\ $$$$\therefore\:\left({sinx}\right)^{{y}} \leqslant\mathrm{1}\:\forall\:{y}\in\mathbb{R} \\ $$$$\therefore\:\mathrm{log}_{{sinx}} \mathrm{2}\:\notin\mathbb{R} \\ $$

Commented by Nimatullah last updated on 22/Mar/22

I don.t know can u explain it for me?

$${I}\:{don}.{t}\:{know}\:{can}\:{u}\:{explain}\:{it}\:{for}\:{me}? \\ $$

Commented by Nimatullah last updated on 22/Mar/22

plz ans

$${plz}\:{ans} \\ $$

Commented by Rasheed.Sindhi last updated on 22/Mar/22

Which step do you not understand?

$${Which}\:{step}\:{do}\:{you}\:{not}\:{understand}? \\ $$

Commented by Nimatullah last updated on 22/Mar/22

what is the soluption of the question can u explain it for what is the valum of x why  it don.t have sulotion   in the last ans i dont know  (sinx)^y ∉R i dont know about it bro

$${what}\:{is}\:{the}\:{soluption}\:{of}\:{the}\:{question}\:{can}\:{u}\:{explain}\:{it}\:{for}\:{what}\:{is}\:{the}\:{valum}\:{of}\:{x}\:{why} \\ $$$${it}\:{don}.{t}\:{have}\:{sulotion}\: \\ $$$${in}\:{the}\:{last}\:{ans}\:{i}\:{dont}\:{know} \\ $$$$\left({sinx}\right)^{{y}} \notin{R}\:{i}\:{dont}\:{know}\:{about}\:{it}\:{bro} \\ $$

Commented by Rasheed.Sindhi last updated on 22/Mar/22

I think I′m wrong. For negative y  (sinx)^y  may be greater than 1.  Thank you Nimatullah to point out  my mistake.

$${I}\:{think}\:{I}'{m}\:{wrong}.\:{For}\:{negative}\:{y} \\ $$$$\left({sinx}\right)^{{y}} \:{may}\:{be}\:{greater}\:{than}\:\mathrm{1}. \\ $$$${Thank}\:{you}\:{Nimatullah}\:{to}\:{point}\:{out} \\ $$$${my}\:{mistake}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com