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A little help on this? (C1)

Given that 3x=9y−13^x = 9^{y - 1}, show that x=2y−2x = 2y - 2.
Reply 1
Avatar for Draconis
Draconis
OP
2
Ah, so since 32=913^2 = 9^1, 32y−2=9y−13^{2y - 2} = 9^{y - 1}?
Reply 2
Avatar for Draconis
Draconis
OP
2
There's no more working out than that? :s-smilie:
Reply 3
Avatar for syathish
syathish
3X=9y-1
Or, 3x=32(y-1)

Here the bases are same therefore there corresponding powers must be equal,
Hence, x=2(y-1)
Or, x=2y-2 (multiplying 2 with all the terms inside the bracket)
paint.jpg17.4KB
i had this in my c1 progress test exam n didnt have a clue where to start
Reply 5
Avatar for nuodai
nuodai
17
Draconis
Given that 3x=9y−13^x = 9^{y - 1}, show that x=2y−2x = 2y - 2.

Because I'm bored, I'll put all the necessary steps you could possibly want :p:
[br]9=32[br]∴3x=32(y−1)[br]∴3x=32y−2[br]∴log⁡33x=log⁡332y−2[br]∴xlog⁡33=(2y−2)log⁡33[br]∴x×1=(2y−2)×1[br]∴x=2y−2\newline[br]9 = 3^2\newline[br] \therefore 3^x = 3^{2(y-1)}\newline[br]\therefore 3^x = 3^{2y-2}\newline[br]\therefore \log_{3} 3^x = \log_{3} 3^{2y - 2} \newline[br]\therefore x\log_{3} 3 = (2y - 2)\log_{3} 3 \newline[br]\therefore x \times 1 = (2y - 2) \times 1 \newline[br]\therefore x = 2y - 2
Reply 6
Avatar for Kevin_B
Kevin_B
1
You don't even need to use logs
What are logs.. that just gets u more confused don't u think
Reply 8
Avatar for Draconis
Draconis
OP
2
No, I've used logs. I'm doing C1 revision and for some reason got stuck on that one.
nuodai
Because I'm bored, I'll put all the necessary steps you could possibly want :p:
[br]9=32[br]∴3x=32(y−1)[br]∴3x=32y−2[br]∴log⁡33x=log⁡332y−2[br]∴xlog⁡33=(2y−2)log⁡33[br]∴x×1=(2y−2)×1[br]∴x=2y−2\newline[br]9 = 3^2\newline[br] \therefore 3^x = 3^{2(y-1)}\newline[br]\therefore 3^x = 3^{2y-2}\newline[br]\therefore \log_{3} 3^x = \log_{3} 3^{2y - 2} \newline[br]\therefore x\log_{3} 3 = (2y - 2)\log_{3} 3 \newline[br]\therefore x \times 1 = (2y - 2) \times 1 \newline[br]\therefore x = 2y - 2


This is ridiculous. I know you are just having a laugh but the effect could be to worry people unnecessarily about their subject knowledge.

For people just getting to grips with Core 1, nuodia's method is no more necessary than adding 45308 and then subtracting 45308 before you start to tackle the problem.
Reply 10
Avatar for nuodai
nuodai
17
Da_hopeful_1
What are logs.. that just gets u more confused don't u think

Forget the logs. All you need to know is that af=ag⇒f=ga^f = a^g \Rightarrow f = g. In other words, if both sides of the equation are powers of the same number (a), then you can just get rid of that number (a) to leave the two functions. The logs are just an intermediate step that you don't need to know until C2 :smile:

So, if 38x+7=382y+638^{x + 7} = 38^{2y + 6}, then x+7=2y+6x + 7 = 2y + 6. This is why you have to know that 9=329 = 3^2, because then you can have both sides of the equation in the form 3something3^{something} and then just get rid of the 3's and bump down the 'somethings'.
nuodai
Forget the logs. All you need to know is that af=ag⇒f=ga^f = a^g \Rightarrow f = g. In other words, if both sides of the equation are powers of the same number (a), then you can just get rid of that number (a) to leave the two functions. The logs are just an intermediate step that you don't need to know until C2 :smile:

So, if 38x+7=382y+638^{x + 7} = 38^{2y + 6}, then x+7=2y+6x + 7 = 2y + 6. This is why you have to know that 9=329 = 3^2, because then you can have both sides of the equation in the form 3something3^{something} and then just get rid of the 3's and bump down the 'somethings'.


Good explanation. You redeemed yourself.

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