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Integration

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Original post by raheem94
For question 26,
23x+1=4x \displaystyle 2^{3x+1}=4^{x}
Take log of both sides,
log(23x+1)=log(4x)    (3x+1)log2=xlog4 \displaystyle log\left(2^{3x+1}\right)=log \left(4^{x}\right) \implies (3x+1)log2=xlog4


So, (3x+1)log2 = xlog4

(3x+1) =xlog4/log2

3x+1=2

3x = 1

x = 1/3

??? which isn't right
Original post by raheem94
log(2x2x3)    log(2x) \displaystyle log\left(\frac{2x^2}{x^3}\right) \implies log\left(\frac2{x}\right)


So what does x = 2?
Reply 62
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'.
Reply 63
Original post by King-Panther
So what does x = 2?


According to the question,
log(2x)=log0.5 \displaystyle log\left(\frac2{x}\right) = log0.5

Remove logs,
2x=0.5 \displaystyle \frac2{x}= 0.5
Reply 64
Original post by just george
24. yes it is true, the first wiki image steve2005 put up states that if f(x)<0 f''(x)<0 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 :smile:


Consider f(x)=x64f(x) = -x^{64}.
Original post by raheem94
According to the question,
log(2x)=log0.5 \displaystyle log\left(\frac2{x}\right) = log0.5

Remove logs,
2x=0.5 \displaystyle \frac2{x}= 0.5


therefore x =2/0.5

x = 1???

I take I substitute value of x into the equation to get log(0.5)
Reply 66
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
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
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.
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
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
Reply 71
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


2x=0.5    2x=12    x×2x=x×12    2=x2    2×2=2×x2    4=x \displaystyle \frac{2}{x} = 0.5 \implies \frac{2}{x} = \frac12 \implies x \times \frac2{x} = x \times \frac12 \implies 2 = \frac{x}{2} \implies 2 \times 2 = 2\times \frac{x}2 \implies 4=x

Your equation was, logx+log2xlogx3=log0.5 \displaystyle logx+log2x-logx^3 =log0.5
log0.5=0.301... \displaystyle log0.5=-0.301...
log4+log2(4)log43=log4+log8log64=0.301... \displaystyle log4+log2(4)-log4^3=log4+log8-log64=-0.301...
Original post by Zuzuzu
Consider f(x)=x64f(x) = -x^{64}.


Unparseable latex formula:

f'(x)=-64x^6^3



so turning point at x=0

Unparseable latex formula:

f''(x)=-4032x^6^2



so at x=0,f(x)=0x=0, f''(x)=0

so not a maximum..?

i dont understand what you're trying to say? is what i said incorrect? :L
Original post by raheem94
2x=0.5    2x=12    x×2x=x×12    2=x2    2×2=2×x2    4=x \displaystyle \frac{2}{x} = 0.5 \implies \frac{2}{x} = \frac12 \implies x \times \frac2{x} = x \times \frac12 \implies 2 = \frac{x}{2} \implies 2 \times 2 = 2\times \frac{x}2 \implies 4=x

Your equation was, logx+log2xlogx3=log0.5 \displaystyle logx+log2x-logx^3 =log0.5
log0.5=0.301... \displaystyle log0.5=-0.301...
log4+log2(4)log43=log4+log8log64=0.301... \displaystyle log4+log2(4)-log4^3=log4+log8-log64=-0.301...


ahhhh

:hugs:
Original post by raheem94
2x=0.5    2x=12    x×2x=x×12    2=x2    2×2=2×x2    4=x \displaystyle \frac{2}{x} = 0.5 \implies \frac{2}{x} = \frac12 \implies x \times \frac2{x} = x \times \frac12 \implies 2 = \frac{x}{2} \implies 2 \times 2 = 2\times \frac{x}2 \implies 4=x

Your equation was, logx+log2xlogx3=log0.5 \displaystyle logx+log2x-logx^3 =log0.5
log0.5=0.301... \displaystyle log0.5=-0.301...
log4+log2(4)log43=log4+log8log64=0.301... \displaystyle log4+log2(4)-log4^3=log4+log8-log64=-0.301...


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
(edited 11 years ago)
and for (26) couldnt you just do?

23x+1=4x 2^{3x+1} = 4^x

23x+1=(2×2)x 2^{3x+1} = (2\times2)^x

23x+1=2x×2x 2^{3x+1} = 2^x\times2^x

23x+1=22x 2^{3x+1} = 2^{2x}

3x+1=2x 3x+1 = 2x

x=1 x = -1

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

23x+1=4x 2^{3x+1} = 4^x

23x+1=(2×2)x 2^{3x+1} = (2\times2)^x

23x+1=2x×2x 2^{3x+1} = 2^x\times2^x

23x+1=22x 2^{3x+1} = 2^{2x}

3x+1=2x 3x+1 = 2x

x=1 x = -1

Guess if your happy with laws of logs and things either way is fine, dont know why i chose to do it this way :tongue:

indeed i could have, thanks.... for 31 is it 8?
Original post by King-Panther
indeed i could have, thanks.... for 31 is it 8?


yes, whats the relevance of your answer though? :smile:

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 32x2 \frac{3}{2}x^2 , so to differentiate it you times it by the power, and then take one off the power?
(edited 11 years ago)
Original post by just george
yes, whats the relevance of your answer though? :smile:


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?
Original post by just george
yes, whats the relevance of your answer though? :smile:

edit:


what do you mean? you have 32x2 \frac{3}{2}x^2 , so to differentiate it you times it by the power, and then take one off the power?


No, the fraction is 32x2 \frac{3}{2x^2}
(edited 11 years ago)

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