Ask your own question, for FREE!
Mathematics 17 Online
OpenStudy (anonymous):

Which of the following is the simplified version of 1/2 log 3 x + 2 log 3 y – 4 log 3 z ?

OpenStudy (anonymous):

\[\frac{ 1 }{ 2 } \log _{3} x+2 \log _{3} y-4\]

hartnn (hartnn):

i assume you'll post choices ? and the expression is \(\large \frac{ 1 }{ 2 } \log _{3} x+2 \log _{3} y-4\log_3z\) right ?

OpenStudy (anonymous):

log 3 sqrt(x) / y^2z^4 log 3 sqrt(xy^2) / z^4 log 3 sqrt(xz^4) / y^2 log 3 x3yz

OpenStudy (anonymous):

\[\log_{3}(\sqrt{x}y^2)/z^4\]

hartnn (hartnn):

ok, we will use these properties, \(a \log b = \log b^a \) \(\log a +\log b = \log ab\) \(\log a -\log b = \log a/b\) heard of them ?

OpenStudy (anonymous):

yes @hartnn

hartnn (hartnn):

so, which one will u use for 1/2 log x ?? (forget about the 3 as of now)

OpenStudy (anonymous):

the middle one ?

hartnn (hartnn):

nopes, \(a \log b =\log b^a\) \((1/2)\log x = .... ?\)

OpenStudy (anonymous):

oh okay

hartnn (hartnn):

i was asking what will 1/2 log x simplify to ?

OpenStudy (anonymous):

log x ^1/2

hartnn (hartnn):

correct! in same way, what about 2 log y =... ?

hartnn (hartnn):

and 4 log z = ... ?

OpenStudy (anonymous):

log y ^2 log z ^4

hartnn (hartnn):

good good! now we will use other properties, log a + log b = log ab log (x^1/2) + log y^2 = ... ?

OpenStudy (anonymous):

log x^1/2y^2

hartnn (hartnn):

correct, log a - log b = log a/b so, log (x^1/2y^2 ) - log z^4 = ... ?

OpenStudy (anonymous):

log 3 (x^1/2y^2) / z^4

hartnn (hartnn):

good work, thats your answer :)

hartnn (hartnn):

las thing, x^1/2 = sqrt x

OpenStudy (anonymous):

thank youu (:

hartnn (hartnn):

welcome ^_^

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!