SOLUTION: If {{{ log( a, x ) = P }}} and {{{ log( a, y ) = 7 }}} . Express {{{ log( a, x^2y ) }}} in terms of P and 7. {{{ log( a, x ) = P }}} => {{{ a^P = x }}} {{{ log( a, y ) = 7 }}} =

Question 559518: If and . Express in terms of P and 7.
=>
=>
Log(a,(a^P)^2a^7 ) => Log( a, a^2Pa^7 )
and I was stuck. I am not even sure of what i'm doing ????
So, please help.

Found 3 solutions by KMST, htmentor, Theo:
Answer by KMST(5328)   (Show Source): You can put this solution on YOUR website!
I think you are trying to go around it the long way to the solution, and getting lost in the woods in between. I would go in there and figure out where you took the wrong fork in the road, but I've learned to be afraid of going through such woods.
I would start from;
= + = +
You would know what to do then.
It is easier to find your way to the center of a labyrinth if you trace the way back from the center. You do not waste your timr and patience bumping into so many dead ends along the way. Less frustration, less errors. (I learned that the hard way).
Answer by htmentor(1343)   (Show Source): You can put this solution on YOUR website!
If and . Express in terms of P and 7.
===================================
For simplicity, we write log_a as log
p = log(x)
7 = log(y)
Using the rules of logarithms, we can write
log(x^2*y) = log(x^2) + log(y) = 2*log(x) + log(y) = 2p + 7

Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
your problem states:
log(a,x) = p
log(a,y) = 7
you want to express log(a,x^2*y) in terms of p and 7.
start with log(a,x^2*y)
this becomes:
log(a,x^2) + log(a,y) which becomes:
2*log(a,x) + log(a,y)
since log(a,x) = p and log(a,y) = 7, your equation becomes:
2*p + 7
it can also be shown as:
2p + 7
the logarithm rules you are using are:
log(x*y) = log(x) + log(y)
log(x^y) = y*log(x)
these rules apply regardless of the base of the logarithm that is being used.

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