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IS ANYONE HERE GOOD… - QuestionCove
OpenStudy (anonymous):

IS ANYONE HERE GOOD AT COLLEGE ALGEBRA AND WILLING TO HELP ME PLEASE

1 year ago
OpenStudy (directrix):

Post the question, okay?

1 year ago
OpenStudy (anonymous):

yes! ill post it right now @Directrix

1 year ago
OpenStudy (anonymous):

|dw:1457821628536:dw| express in therms of logarithm of x,y, or z

1 year ago
OpenStudy (jhonyy9):

you need using log a/b = log a -log b log a*b = log a + log b hope this will help you

1 year ago
OpenStudy (directrix):

Applying the first log law from above: log y^(1/2) - log ( (x^6 * z^ (1/3) ) ) = Now, the second law from above: log y^(1/2) - [ log ( (x^6 ) + log z^ (1/3) ] = (1/2) log y - 6 log x - (1/3) log z -------- Review Laws of Logs (See attachment)

1 year ago
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