Anee_rk
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#1
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#1
Solve the following simultaneous logarithmic equations
y^logx=100
log(sqrt(xy/10))=1
given further that x , y ∈ℝ , with x > 0 , y > 0.
Can someone help me with this question, I keep getting an unsolvable quadratic.
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mqb2766
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#2
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#2
(Original post by Anee_rk)
Solve the following simultaneous logarithmic equations
y^logx=100
log(sqrt(xy/10))=1
given further that x , y ∈ℝ , with x > 0 , y > 0.
Can someone help me with this question, I keep getting an unsolvable quadratic.
upload what you tried?
Im presuming log is tne natural log, but to get a handle on (guess) the solution, you could assume it was base 10.
Last edited by mqb2766; 1 month ago
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Anee_rk
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#3
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#3
(Original post by mqb2766)
upload what you tried?
Im presuming log is tne natural log, but to get a handle on (guess) the solution, you could assume it was base 10
I realised i made a simple mistake and figured it out, but thanks for the help. Though i am confused on the next one, could i upload that instead?
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mqb2766
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#4
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#4
(Original post by Anee_rk)
I realised i made a simple mistake and figured it out, but thanks for the help. Though i am confused on the next one, could i upload that instead?
sure
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Anee_rk
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#5
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#5
(Original post by mqb2766)
sure
Name:  madasmaths q3 (2).png
Views: 11
Size:  59.6 KB
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mqb2766
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#6
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#6
(Original post by Anee_rk)
Name:  madasmaths q3 (2).png
Views: 11
Size:  59.6 KB
And what do you understand / have problem with?
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Anee_rk
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#7
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#7
(Original post by mqb2766)
And what do you understand / have problem with?
im not really sure where to start
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Anee_rk
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#8
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#8
(Original post by Anee_rk)
im not really sure where to start
Does the first line mean its a constant increasing/decreasing function? how does that help me?
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mqb2766
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#9
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#9
(Original post by Anee_rk)
im not really sure where to start
The first line says the rate of change of the gradient is constant.
So write that down clearly and try and solve for that.
Last edited by mqb2766; 1 month ago
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Anee_rk
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#10
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#10
(Original post by mqb2766)
The first line says the rate of change of the gradient is constant.
So write that down clearly and try and solve for that.
ive got a few simultaneous equations:
m+c=-1/2
2=m/2+c+d
9m-3c-d=0
Do i solve for m,c, and d, then equate the final function to y=2x to work out where they intersect and then use definite integration to find out area under the curve and subtract it from the area under the line?
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mqb2766
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#11
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#11
(Original post by Anee_rk)
ive got a few simultaneous equations:
m+c=-1/2
2=m/2+c+d
9m-3c-d=0
Do i solve for m,c, and d, then equate the final function to y=2x to work out where they intersect and then use definite integration to find out area under the curve and subtract it from the area under the line?
What curve did you get from the first part (what are the parameters m,c,d)?

Assuming its a quadratic etc, then yes you need to find the points of intersection (roots of curve-line) then use integration to find the area between the curve and the line. So integrate curve-line between the two roots.
Last edited by mqb2766; 1 month ago
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Anee_rk
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#12
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#12
(Original post by mqb2766)
What curve did you get from the first part (what are the parameters m,c,d)?

Assuming its a quadratic etc, then yes you need to find the points of intersection (roots of curve-line) then use integration to find the area between the curve and the line. So integrate curve-line between the two roots.
Sorry i had to redo my workings because i made some mistakes but hopefully it is right this time. I got:
m=-1 c=0.5 d=1.5
But yes it is a quadratic.
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mqb2766
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#13
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#13
(Original post by Anee_rk)
Sorry i had to redo my workings because i made some mistakes but hopefully it is right this time. I got:
m=-1 c=0.5 d=1.5
But yes it is a quadratic.
Ive not done the numbers, but a quadratic is easy to verify for the 3 points.
So just integrate the curve-line between the roots.
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Anee_rk
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#14
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#14
(Original post by mqb2766)
Ive not done the numbers, but a quadratic is easy to verify for the 3 points.
So just integrate the curve-line between the roots.
okay thanks!
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Anee_rk
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#15
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#15
Name:  mad as maths q1 (2).png
Views: 4
Size:  14.0 KBWhat is the easiest way to tackle this question?
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mqb2766
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#16
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#16
(Original post by Anee_rk)
Name:  mad as maths q1 (2).png
Views: 4
Size:  14.0 KBWhat is the easiest way to tackle this question?
Any ideas?
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Anee_rk
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#17
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#17
(Original post by mqb2766)
Any ideas?
I see, i'll start by factorising the quadratic
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Mathaddict100
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#18
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#18
(Original post by Anee_rk)
Solve the following simultaneous logarithmic equations
y^logx=100
log(sqrt(xy/10))=1
given further that x , y ∈ℝ , with x > 0 , y > 0.
Can someone help me with this question, I keep getting an unsolvable quadratic.
I dont know
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mqb2766
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#19
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#19
(Original post by Anee_rk)
I see, i'll start by factorising the quadratic
Thats one way, then expand each factor^7 up to quadratic then multply out.
Or write it as (1 + (3x+2x^2))^7 and write out the first three terms of the binomial in (3x+2x^2) and combine.
Not a great difference in either way.
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the bear
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#20
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#20
(Original post by mqb2766)
upload what you tried?
Im presuming log is tne natural log, but to get a handle on (guess) the solution, you could assume it was base 10.
because the numbers 10 and 100 feature prominently, it is likely that the logarithm is base 10 :beard:
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