Algebra

just renenber the form of logxY=z iss hte same thing as x^z=Y.

sow yore queton shuld revsolve itself as 4^x=32x+16y

doun. it is duun. tadsah its cuomplekly don

Mayra:

log(4)(32x+16y)

Hi guest,
This line in your answer is definitely correct
"just renenber the form of logxY=z iss hte same thing as x^z=Y."

Here you have assumed that the 4 was intended as a base.
So what you have here is
log ₄A=Z where A=(32x+16y) (I have just used A and Z so it is not going to be confused with the original x and y)

rearranging this we get
4 ^Z=A

4 ^log₄A= (32x+16y)

A = 32x+16y which isn't very helpful because we already knew that.
--------------------------------------------------------------------------------------------
Hi Mayra,
I always keep this page handy when I am anwering log questions
http://en.wikipedia.org/wiki/List_of_logarithmic_identities

If this was the intended question
log ₄(32x+16y)
then
=log ₄(2x+y)16
=log ₄(2x+y) + log ₄16
=log ₄(2x+y) + log ₄4 ²
=log ₄(2x+y) +2 log ₄4
=2 + log ₄(2x+y)
-----------------------------------------------------------------------------------------------------
Hi Mayra,
However maybe the original question was meant to be log[4(32x+16y)] base 10 or whatever
=log[4*16(2x+y)]
=log[4*16]+log(2x+y)
=log 2 ⁶ + log(2x+y)
=6log2 + log(2x+y)
I don't really think that this is more simple but not to worry.
-----------------------------------------------------------------------------------------------------