Logarithms C2 help

Basically what it's saying that 8 can be written as 2 to the power of 3, and 4 can be written as 2 to the power of 2. They are now both of the same base (2, that is) so any power on either side must be the same too (as they are simultaneous equations) You just multiply out the brackets of the powers and get rid of the base 2.

how did the 2s disappear though

They are the same bases so they disappear.

x^y=x^z

then y=z

how do they cancel out

They cancel out because both sides must be equal.

In your question both sides involve 2 to the power of something. Seeing as 2=2 then the powers must also be equal (because both sides are equal) leaving you with just the powers of one side being equal to the powers of the other side (as they must have the same effect on 2, because, again the two sides are equal).

Sounds like im repeating myself a lot, i know, but its hard not to- theres no other way really to explain this to you as its just a single simple step. I'm sure it'll make sense to you if you just think a little more on it.

If for example the last line but 1 had been
$4^{3y}=2^{2(2x+3)}$ then you couldn't have simply canceled 4 and 2 out to give
$3y=4x+6$ Instead, you would have to write 4 in powers of 2
$(2^2)$ so your solution would have looked like:
$(2^2)^{3y}=2^{2(2x+3)} [br] (2)^{6y}=2^{2(2x+3)}$
and THEN you could cancel the bases on either side (seeing as they are now both two) to give $6y=4x+6.$

Hope this clarifies your problem a bit...

EDIT: SORRY BEAR WITH ME- FIRST TIME USING LATEX AND HAVING SOME PROBLEMS

thank you but could you explain how they cancel

Check "Logarithmic identities" in the wikipedia article about "Logarithm". The power rule especially.

8^y = 4^(2x+3) <=>
2^(3y) = 2^(4x+6) <=>
ln(2^(3y)) = ln(2^(4x+6)) <=> (use the power rule for logarithms in next step)
3y * ln(2) = 4x+6 * ln(2) <=>
3y = 4x+6