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Mathematics 14 Online
OpenStudy (anonymous):

Is this equation true? please help!

OpenStudy (anonymous):

What equation?

OpenStudy (anonymous):

\[4\log_{3} x+\frac{ 1 }{ 2 }\log_{3} y-2\log_{3}2=\log_{3}\frac{ x^4\sqrt{y} }{ z^2 } \]

OpenStudy (anonymous):

sorry not my area of expertise, so to speak.

OpenStudy (anonymous):

Is the 2 before the equal sign supposed to be a z?

OpenStudy (anonymous):

These are all of the same base. So remember: log A + log B - log C is equal to log(ABC)

OpenStudy (anonymous):

Also, you can move the constants back into the log using: n log x = log x^n

OpenStudy (anonymous):

yes it´s a z sorry

OpenStudy (anonymous):

log(ab/c)*

OpenStudy (anonymous):

@CalebBeavers Oh yeah, whoops. log(AB/C)

OpenStudy (anonymous):

Yep, so since that was supposed to be a z then the equation is true.

OpenStudy (anonymous):

is bc of the way muntoo explained it?

OpenStudy (anonymous):

Yes, so: \[\Large {4 \log_3 x + {1\over2} \log_3 y -2\log_3 z \\ = \log_3 x^4 + \log_3 \sqrt{y} - \log_3 z^2 \\ = \log_3 ({x^4\sqrt{y}\over z^2}) }\] Using my rules above.

OpenStudy (anonymous):

thanks! what properties did you use to find that out?

OpenStudy (anonymous):

In the second line, you move the constants in. n log x = log x^n In the third line, you group the logs together. log A + log B - log C = log (AB/C) Note that we are talking about logs of the same base, in this case, 3.

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