Re: law of indices problem

Hey CuriousChris.

The basic reason is that 3^2 = 9 and power laws work in that (x^a)^b = x^(ab) so if you let x = 3 a = 2 and b = x+1 then (3^2)^(x+1) = 3^(2(x+1)) = 3^(2x+1) = 9^(x+1).

So if 3^(2x+2)/3^(2x-1) using exponent laws if you divide you subtract the bottom index from the top to get 3^(2x+2 - (2x-1)) = 3^(2+1) = 3^3.

Re: law of indices problem

In problems like this you have to make the bases the same so the 9 becomes 3^2 giving

3^(2x + 2) / 3^(2x - 1) then like chiro says using the second law of indices division become subtraction (when the bases match) giving

3^(2x + 2) - (2x - 1) = 3^(2 + 1) = 3^3 = 27.

Good luck.

Re: law of indices problem

Hi Guys this is my third attempt to reply. Sorry I keep getting dragged away.

Thanks for your invaluable assistance.

I will remind my son when dealing with such problems to try to get all bases the same if possible. Then to think of the laws of indices and see which apply.

Thanks again. you'll probably see a lot more of me over the next year as my son does year 12 :)

CC

Re: law of indices problem

We welcome him or you any time at all. (Nod)

-Dan