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1. 8). Solve the equation

I changed the base and got to:

I tried to use the power rule remove the coefficient of 2 on the LHS and raise the log on the LHS to 2, but according to wolfram that gives a different answer.

Without Power Rule
With Power Rule
2. At your stage, you really oughtn't be using Wolfram. However, you've inputted the expression incorrectly (the way you've put it Wolfram thinks you mean log * log = log^2, rather than log(whatever^2):

http://www.wolframalpha.com/input/?i...5E2-19x%2B2%29
3. Ahh ok thanks. What should I be trying to do from this point though? Am I able to solve the quadratic, or expand the (2-3x)^2 on the LHS? Or do the logs stop me from doing that?
4. 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.
5. hmmm
6. (Original post by ViralRiver)
8). Solve the equation

I changed the base and got to:

I tried to use the power rule remove the coefficient of 2 on the LHS and raise the log on the LHS to 2, but according to wolfram that gives a different answer.

Without Power Rule
With Power Rule

Change the log_9() base to a base of log 3()
so you get
7. (Original post by Demonic Entity)
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.
yeah
8. (Original post by Demonic Entity)
yeah
9. If you have log(a) = log(b), you can conclude that a = b (if they're in the same base).

i.e. (2-3x)^2 = 6x^2 - 19x + 2.

See, these questions aren't so hard, are they?
10. Oh, that makes more sense. So is this true:

EDIT: Hmm, I see you used the power rule, does that mean that that if then even though the indices are different?
11. I'm stuck on the mixed exercise as well >_<

question no. 7
Find the values of x for which log_3x-2log_x3=1

this is as far I have got into in number 7, my workings is as follows:

log_3x-2log_x3=1
log_3x+2/log_x3=1
let log_3x=y
y+2/y=1
y^2+2=y
y^2-y+2=0
(y-1)(y+2)=0 or (y-1)(y-2)=0?? <--- this is the bit I'm stuck on, you gotta factorise here, but either way (y-1) & (y+2) together or (y-1) & (y-2) together added together won't give -y, it will only give 2 or -3.

--------------------------

I am also stuck on question 12.

a) Given that 3+2log_2x=log_2y, show that y=8x^2.
b) Hence, or otherwise, find the roots α and β, where α < β, of the equation 3+2log_2x=log_2(14x-3).
c)Show that log_2α=-2
d) Calculate log_2β,giving youranswer to 3 significant figures.

Help may someone help me, your help will be greatly appreciated.
12. I'd appreciate it if you kept your questions in a separate topic . Although I did complete question 7:

or

13. (Original post by ViralRiver)
Oh, that makes more sense. So is this true:

EDIT: Hmm, I see you used the power rule, does that mean that that if then even though the indices are different?

Then yes, you can conclude that b^c = d^e. Why? Just let f = b^c and let g = d^e, then

14. Sorry, what I meant was, if then ?
15. (Original post by ViralRiver)
Sorry, what I meant was, if then ?
No, because

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