Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 88663 by mary_ last updated on 12/Apr/20

$$\{\begin{array}{l}{log}_{2}x+{log}_{4}y=4\\ x.y=8\end{array}$$

Answered by jagoll last updated on 28/Apr/20

$${\log }_2\left(\mathrm{x}\right)+\frac{1}{2}{\log }_2\left(\frac{8}{\mathrm{x}}\right)=4$$$${\log }_2\left(x\right)+{\log }_2\sqrt{\frac{8}{x}}={\log }_2\left(16\right)$$$$x\sqrt{\frac{8}{x}}=16\Rightarrow \sqrt{8x}=\sqrt{256}$$$$x=\frac{256}{8}=32soy=\frac{8}{32}=4^{-1}$$

Commented by john santu last updated on 12/Apr/20

$$correctionsir$$$$itshouldbe$$$${\log }_2\left(x\right)+{\log }_2\sqrt{\frac{8}{x}}=4$$$$\Rightarrow x\sqrt{\frac{8}{x}}=16\Rightarrow \sqrt{8x}=16$$$$x=\frac{16.16}{8}=32.theny=\frac{8}{32}=\frac{1}{4}$$

Commented by jagoll last updated on 12/Apr/20

$$ooyes.thankyoumrjohn$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com