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 33103 by mondodotto@gmail.com last updated on 10/Apr/18

Commented by Rasheed.Sindhi last updated on 10/Apr/18

x=y=z=0 In this case y≠((2xz)/(x+z))

x=y=z=0Inthiscasey≠2xzx+z

Answered by MJS last updated on 10/Apr/18

1.: b^2 =ac ⇒ a=(b^2 /c) 2.: b^y =c^z ⇒ c=b^(y/z) 3. (2. in 1.): a=(b^2 /b^(y/z) )=b^(2−(y/z)) =b^((2z−y)/z) 4.: a^x =b^y ⇒ y=x((ln a)/(ln b)) 3. in 4.: y=x((ln b^((2z−y)/z) )/(ln b)) y=x(((((2z−y)/z))ln b)/(ln b)) y=((2xz−xy)/z) y=((2xz)/(x+z))

1.:b2=ac⇒a=b2c2.:by=cz⇒c=byz3.(2.in1.):a=b2byz=b2−yz=b2z−yz4.:ax=by⇒y=xlnalnb3.in4.:y=xlnb2z−yzlnby=x(2z−yz)lnblnby=2xz−xyzy=2xzx+z

Terms of Service

Privacy Policy

Contact: info@tinkutara.com