SOLUTION: if r=log*base4* x and s=log*base4* y how can i write log*base4*(64x^(6)y^(3))^(2) using the variables please help im stuck. ive tried 4096r^(12)+s^(6) and that doesnt make sense

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: if r=log*base4* x and s=log*base4* y how can i write log*base4*(64x^(6)y^(3))^(2) using the variables please help im stuck. ive tried 4096r^(12)+s^(6) and that doesnt make sense      Log On


   

Question 366281: if r=log*base4* x and s=log*base4* y
how can i write log*base4*(64x^(6)y^(3))^(2) using the variables please help im stuck. ive tried 4096r^(12)+s^(6) and that doesnt make sense
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
r=log%284%2C+%28x%29%29
s=log%284%2C+%28y%29%29
log%284%2C+%28%2864x%5E6y%5E3%29%5E2%29%29
What we will be doing is using the properties of logarithms:
  • log%28a%2C+%28p%2Aq%29%29+=+log%28a%2C+%28p%29%29+%2B+log%28a%2C+%28q%29%29
  • log%28a%2C+%28p%2Fq%29%29+=+log%28a%2C+%28p%29%29+-+log%28a%2C+%28q%29%29
  • log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28q%29%29

to break apart the argument of your logarithm. The reason we do this is that we want to find an equivalent expression written in terms of log%284%2C+%28x%29%29 and log%284%2C+%28y%29%29.
We will start by using the third property to move the exponent of 2 out in front as a coefficient:
2%2Alog%284%2C+%2864x%5E6y%5E3%29%29
Next we will use the first property (because the argument is a product) to split the logarithm into an equivalent sum:
2%2A%28log%284%2C+%2864%29%29+%2B+log%284%2C+%28x%5E6%29%29+%2B+log%284%2C+%28y%5E3%29%29%29
Note the use of parentheses! I replaced the single logarithm with a sum of three logarithms and when I did so I put the sum in parentheses. When you make substitutions like this, it is an extremely good habit to surround your replacement expression in parentheses. In this case it helps us know that the 2 in front should be distributed to all the logarithms and not just the first logarithm.
Next, on the last two logarithms, we will use the third property again to move the exponenets out in front:
2%2A%28log%284%2C+%2864%29%29+%2B+6%2Alog%284%2C+%28x%29%29+%2B+3%2Alog%284%2C+%28y%29%29%29
The last two logarithms, we were told, are "r" and "s" respectively. And the first logarithm is a 3. (Since 4%5E3+=+64, log%284%2C+%2864%29%29+=+3). Substituting these into our expression we get:
2*((3) + 6*(r) + 3*(s)) (Again with the parentheses!)
which simplifies as follows:
2(3 + 6r + 3s)
and with the Distributive Property:
6 + 12r + 6s
or
12r + 6s + 6