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1. (Original post by raheem94)
For question 26,

Take log of both sides,
So, (3x+1)log2 = xlog4

(3x+1) =xlog4/log2

3x+1=2

3x = 1

x = 1/3

??? which isn't right
2. (Original post by raheem94)
So what does x = 2?
3. (Original post by King-Panther)
So, (3x+1)log2 = xlog4

(3x+1) =xlog4/log2

3x+1=2

3x = 1

x = 1/3

??? which isn't right
(3x+1) =xlog4/log2
3x+1=2x

You missed out the 'x'.
4. (Original post by King-Panther)
So what does x = 2?
According to the question,

Remove logs,
5. (Original post by just george)
24. yes it is true, the first wiki image steve2005 put up states that if then f has a local maximum at x.
i.e. if the second derivative of f(x) is less than 0 (negative), there is a local maximum.
So clearly all local maximum points will produce a negative second derivative
Consider .
6. (Original post by raheem94)
According to the question,

Remove logs,
therefore x =2/0.5

x = 1???

I take I substitute value of x into the equation to get log(0.5)
7. (Original post by King-Panther)
therefore x =2/0.5

x = 1???

I take I substitute value of x into the equation to get log(0.5)
2/0.5=4 not 1
8. (Original post by raheem94)
2/0.5=4 not 1
2/x = 0.5, if you make x the subject, you multiply 2 by .5 ....

When I sub 4 into the equation logx+log2x+logx^3 i dont get 0.5
9. (Original post by King-Panther)
2/x = 0.5, if you make x the subject, you multiply 2 by .5 ....
No, you'd multiply through by x, then divide through by 0.5, so it does equal 4.
10. (Original post by Brit_Miller)
No, you'd multiply through by x, then divide through by 0.5, so it does equal 4.
of course, however, when i sub 4 into the equation i dont get 0.5
11. (Original post by Brit_Miller)
No, you'd multiply through by x, then divide through by 0.5, so it does equal 4.
for Q)31, i got 8
12. (Original post by King-Panther)
2/x = 0.5, if you make x the subject, you multiply 2 by .5 ....

When I sub 4 into the equation logx+log2x+logx^3 i dont get 0.5

13. (Original post by Zuzuzu)
Consider .

so turning point at x=0

so at

so not a maximum..?

i dont understand what you're trying to say? is what i said incorrect? :L
14. (Original post by raheem94)

ahhhh

15. (Original post by raheem94)

Q)31 i got 8

also, for 33, what do i do when 3/2x^2, do i multiply 3 by 2x^2 do its 6x^-2
16. and for (26) couldnt you just do?

Guess if your happy with laws of logs and things either way is fine, dont know why i chose to do it this way
17. (Original post by just george)
and for (26) couldnt you just do?

Guess if your happy with laws of logs and things either way is fine, dont know why i chose to do it this way
indeed i could have, thanks.... for 31 is it 8?
18. (Original post by King-Panther)
indeed i could have, thanks.... for 31 is it 8?

edit:
(Original post by King-Panther)
also, for 33, what do i do when 3/2x^2, do i multiply 3 by 2x^2 do its 6x^-2
what do you mean? you have , so to differentiate it you times it by the power, and then take one off the power?
19. (Original post by just george)
Dy/dx tells us what the gradient is when we at x when we sub it in, d2y/dx2 tells us in which direction its going, its a positive number so its a minimum going in the positive direction?
20. (Original post by just george)

edit:

what do you mean? you have , so to differentiate it you times it by the power, and then take one off the power?
No, the fraction is

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