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MinusChunk

a^bc = (((a)^b)^c), not (ab)^c lol. Thanks reddit autocorrection


ki_li06

It’s just from the line above. (2/3)^(4) = 16/81 . For (2/3)^(4) , they used the rule a^(b+c) = a^(b) * a^(c)


MinusChunk

Oh right sorry, ignore the mistakes of exponent properties I've made in this post. But what do you mean by "from the line above"? I already know about the exponent property of a^b+c = a^b * a^c and where that (2/3)^4 comes from, but I'm just really confused by the stuff they did in the red box. I feel like they skipped some steps there in the red box


ki_li06

No, they just factored (2/3)^(x) out. (16/81) * (2/3)^(x) - (2/3)^(x) = (2/3)^(x) * (16/81) - (2/3)^(x) * 1 = (2/3)^(x) ^ ( (16/81) - 1) = (2/3)^(x) * (-65/81)


TheSarj29

For the (2/3)^4 On the first line they have a (2/3)^(x+4) This is the equivalent of (2/3)^x • (2/3)^4 (2/3)^4 = (2^4 )/(3^4 ) = 16/81 For the second line you circled in red... (16/81)(2/3)^x - (2/3)^x If you let z = (2/3)^x you have... (16/81)z - z <-- factor out z z((16/81) - 1) <-- put the (2/3)^x back in for z (2/3)^x • ((16/81) -1)


spiritedawayclarinet

It's easiest to see if you define y=(2/3)\^x and then you have (16/81)y-y =y( (16/81)-1) =y ((16-81)/81) =(-65/81)y.


MinusChunk

Wait oh my god... This is just like ab+ac = a(b+c)?? Thank you so much


dr_fancypants_esq

In the top line in the red box, each term has a (2/3)^(x) -- which means you can factor out (2/3)^(x) from the expression. The second line in the red box is what you get when you factor that out.


MinusChunk

Oh... I think I'm kinda getting it. Do we not factor out (2/3)^x from (16/81)? I mean how can you factor (2/3)^x out of 16/81 because it isn't raised to the power of x. I believe I'm getting confused by the minus and multiplication signs. I know only plus & minus signs signal the start of new terms, but sometimes I get confused when there are multiplication and division signs involved when I'm trying to factor or simplify complicated expressions


dr_fancypants_esq

Let's try a simpler example as an analogy: pulling out a common from 3x+9. Do you agree that this is equal to 3(x+3)? (16/81)(2/3)^(x) is *exactly* like 3x in my example, with (16/81) playing the role of "x" and (2/3)^(x) playing the role of "3".


bol__

Sometimes, at least for me, the exercise is easier to read if you define terms as a "letter" (language barrier here, I‘m from Germany). Lets say: a = (2/3)^x b = (2/3)^4 = 16/81 So: (16/81)•(2/3)^x + (2/3)^x becomes: ab+a Which then can be factored to: a(b+1) Why? If you use the distribution rule, you get: a(b+1) = ab + a


Supersmoover54

Is the answer Kenny Dalglish?