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Mathematics 20 Online
OpenStudy (anonymous):

if log (x+y)/7=1/2(logx+logy) prove that x/y+y/x=47

hartnn (hartnn):

2 log [(x+y)/7] = log x + log y you know logarithm properties ?

hartnn (hartnn):

how will you simplify log x + log y ?

hartnn (hartnn):

no? / yes? / just give me answer?

OpenStudy (anonymous):

hi are you speaking to me?

hartnn (hartnn):

hey dane, yes :) since you asked that question, i wanted to know whether you know some logarithm properties which will help us solve this problem :)

OpenStudy (anonymous):

cool !!!:) yes i know the logarithm properties

hartnn (hartnn):

nice , how will you simplify log x + log y ?

OpenStudy (anonymous):

log xy?

hartnn (hartnn):

correct! now \(\log a^b = b \log a\) so if i had [b log a], i could write it as \(\log a^b\) right ? similarly how will i write 2 log [(x+y)/7] as ?

OpenStudy (anonymous):

log(x+y/7)^2

hartnn (hartnn):

oh wow, you're good at logarithms :) \(\log [(x+y)/7]^2 = \log xy\) so we can simply write \( [(x+y)/7]^2 = xy\) isn't it ? can you simplify that further ?

hartnn (hartnn):

expand the left side

hartnn (hartnn):

everything after this is purely algebraic :)

OpenStudy (anonymous):

can you help me with the key to type square on the keyboard instead of typing ^

OpenStudy (anonymous):

thanks a million god bless you

OpenStudy (anonymous):

you gone?

hartnn (hartnn):

just went to help other users too :) and you're most welcome ^_^ type \(x^2 \) as `\(x^2 \)`

OpenStudy (anonymous):

so kind of you !! can i ask another?

hartnn (hartnn):

sure :) ask as many as you want :)

hartnn (hartnn):

since you're new here, \(\Huge \mathcal{\text{Welcome To OpenStudy}\ddot\smile} \)

OpenStudy (usukidoll):

@hartnn !!! yay

OpenStudy (anonymous):

thanks

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