Question

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

Answer 1

Post the question, okay?

Answer 2

yes! ill post it right now @Directrix

Answer 3

express in therms of logarithm of x,y, or z

Answer 4

you need using

log a/b = log a -log b
log a*b = log a + log b

hope this will help you

Answer 5

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

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Review Laws of Logs (See attachment)