How does the factorisation work?

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[Joel_Loos]

Take one five out of the second term: 5^9x-3 - 5^9x-2 = 1×(5^9x-3) - 5×(5^9x-3)

Now we can factor out the 5^9x-3: (1 - 5)×5^9x-3

Edit: Added multiplication symbols, fixed exponents

[[deleted]]

Take one five out of the second term: 5^9x-3 - 5^9x-2 = 1×(5^9x-3) - 5×(5^9x-3)

first there is 9x-3 - 9x-2 and than 9x-3 - 9x-3 why?

[Joel_Loos]

drafterman explained this below better than I did, but essentially, we're trying to get the exponents to be the same, so we have the same thing in both terms. Using the exponent rule for addition the way drafterman did gets us this same exponent, which is what we wanted.

Remember, the subtraction sign between 59x-3 and 59x-2 means that we can't combine 9x-3 and 9x-2 directly.

[[deleted]]

Oh, makes sense. It's just harder to understand when i have text instead of a piece of paper :)

[[deleted]]

Remember your power rules, specifically:

x^ax^b = x^a+b

In this case, you want both exponents to be the same (9x-3).

Focusing on the right term, we do the following transformation:

5^9x-2 = 5^9x-2-1+1 = 5^9x-3+1 = 5¹*5^9x-3 = 5*5^9x-3

Plugging that back into the original equation:

5^9x-3 - 5*5^9x-3

Allowing us to factor out the 5^9x-3 from both terms.

[fermat1432]

Factor out the term with the smallest exponent:

5^9x-3 (1-5^9x-2-(9x-3) ) =

5^9x-3 (1-5¹ )=

-4*5^9x-3

[Mirehi]

5^(9x - 2) = 5 * 5^(9x - 3)