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How do you solve this? simple terms please :(

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what's an easy way to solve a question like this when the bottom numbers aren't the same?

many thanks
Original post by vxdturbo12
WP_20151118_010.jpg

what's an easy way to solve a question like this when the bottom numbers aren't the same?

many thanks


Find the prime factorisation of each number.
So what you do it first you add the powers and times the number so it would look like this
36^6/24^12
Then you divide the numbers and subtract the powers
And it will look like this 1.5^-6
This is how I was taught how to do it sorry if it's wrong
Reply 3
Original post by morgan8002
Find the prime factorisation of each number.


what do you mean exactly? :/
Reply 4
Original post by vxdturbo12
what do you mean exactly? :/


A number can be broken down into a product of prime numbers

9 = 3 * 3
16 = 2 * 2 * 2 * 2
35 = 5*7

etc...
Original post by vxdturbo12
what do you mean exactly? :/


As Zacken said, each positive integer(whole number) can be uniquely factorised into its prime factors.
For example 30 = (2)(3)(5).

If you write the numerator and denominator in their prime factorisations, you can cancel their like factors to write the fraction in its simplest form.
Reply 6
Original post by morgan8002
As Zacken said, each positive integer(whole number) can be uniquely factorised into its prime factors.
For example 30 = (2)(3)(5).

If you write the numerator and denominator in their prime factorisations, you can cancel their like factors to write the fraction in its simplest form.


like this? If so what's next :/

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Original post by vxdturbo12
like this? If so what's next :/

WP_20151120_028.jpg


Use index laws and then cancel like factors.
Reply 8
Original post by morgan8002
Use index laws and then cancel like factors.


I'm not 100 percent sure how to because I've got an index law sheet in front of me but none say what to do if you have brackets or what to do, any advice?

thank you btw
(ab)cabc(a^b)^c\equiv a^{bc}

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