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# Quick question for logs. watch

1. I got a question im stuck on atm.

4(3^2x+1) + 17(3x) - 7.

I let y = 3x. But somehow got muddled up with the first part of the substition.

Any help appreciated.
2. If y = 3^x
then y^2 = 3^2x

The 4(3^2x+1) can be broken down into:
4(3^2x * 3)

Which then becomes:
4(y^2 x 3)
--> 4(3y^2)
---> 12y^2

Hope that helps. Quote me if you need help with the 2nd bit, or need help understanding something! (Logs are the only things I can do in C2 xD)
3. hey i just did that question and it wont factorise
did the formula and both times y is less than 0 so does that mean you can't get an answer or have i done it wrong?
i got (3y-1)(4y+7)

so y can only equal 1/3 in this case as negatives don't count.
5. (Original post by reggaeinyourjeggae)
hey i just did that question and it wont factorise
did the formula and both times y is less than 0 so does that mean you can't get an answer or have i done it wrong?
I had the same probleymo, i'll have a look at it again using info the above user gave.
6. (Original post by CityOfMyHeart)
If y = 3^x
then y^2 = 3^2x

The 4(3^2x+1) can be broken down into:
4(3^2x * 3)

Which then becomes:
4(y^2 x 3)
--> 4(3y^2)
---> 12y^2

Hope that helps. Quote me if you need help with the 2nd bit, or need help understanding something! (Logs are the only things I can do in C2 xD)
thanks, it was one of the things i thought about doing but actually didnt do, apprecitated.
7. (Original post by puma21)
i got (3y-1)(4y+7)

so y can only equal 1/3 in this case as negatives don't count.
ahh of course,man im stupid
cheers
i should probly get back to the start of c1 and learn to factorise stuff..

edit:you said use the quadratic formula? that didnt work ;s but simple factorisation did.. :s
ohwell
8. Did you get a negative in the square root bit you are suppose to add due to the double negative, if so i made the same mistake first time around.
9. no, i just looked back and forgot to multiply ac by 4...:\
atleast its just stupid mistakes and not fundemental misunderstanding

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