Help with a math problem?
- Thread starter cheesyPOOF5
- Start date
- Views 2,297
- Replies 9
- Joined
- Dec 5, 2005
- Messages
- 2,230
- Trophies
- 1
- Age
- 30
- Location
- Richmond, VA
- Website
- Visit site
- XP
- 599
- Country
joemommasfat's method is correct but I suggest you learn the "power laws" as that is what this seems to want to teach.
2^2*2^2=2^4 aka a^x.a^y = a^(x+y)
You might like to think of this for the original purpose it was designed for.
Mathematicians and everyone really is lazy so if something can be written in shorthand it will, not to mention a large group of letters can be very confusing at a glance.
so a.a is written as a^2
a.a.a is written as a^3
a.a.a.a is written as a^4
divided by you take instead of adding a^x/a^y = a^(x-y) (instead you might like to think of it by doing the old method of if you have a divide put it 1/number and multiply it instead and when you know 1/a = a^-1 is becomes a bit easier)
square numbers are inverse powers
i.e. square root of 2 is 2^1/2
power on power gets multiplied
(2^2)^2
=2^4
hope you get it sorted.
2^2*2^2=2^4 aka a^x.a^y = a^(x+y)
You might like to think of this for the original purpose it was designed for.
Mathematicians and everyone really is lazy so if something can be written in shorthand it will, not to mention a large group of letters can be very confusing at a glance.
so a.a is written as a^2
a.a.a is written as a^3
a.a.a.a is written as a^4
divided by you take instead of adding a^x/a^y = a^(x-y) (instead you might like to think of it by doing the old method of if you have a divide put it 1/number and multiply it instead and when you know 1/a = a^-1 is becomes a bit easier)
square numbers are inverse powers
i.e. square root of 2 is 2^1/2
power on power gets multiplied
(2^2)^2
=2^4
hope you get it sorted.
Similar threads
