Solve the equation with the help of determinants
$begingroup$
The following set of equations is given:
$$begin{cases}{{x^2z^3over y}=e^8\{y^2zover x}=e^4\{x^2yover z^4}=1}end{cases}$$
I have solved problems before with three variables in three equations using determinant method, that is Cramer's rule. But in this problem the variables are in multiplication. I seem to have no idea on how to solve this. Please help ASAP.
matrices determinant
edited Jan 15 at 9:22
Mostafa Ayaz
15.5k3939
asked Jan 15 at 7:58
Vibhor KumarVibhor Kumar
61
$endgroup$
add a comment |
$begingroup$
The following set of equations is given:
$$begin{cases}{{x^2z^3over y}=e^8\{y^2zover x}=e^4\{x^2yover z^4}=1}end{cases}$$
I have solved problems before with three variables in three equations using determinant method, that is Cramer's rule. But in this problem the variables are in multiplication. I seem to have no idea on how to solve this. Please help ASAP.
matrices determinant
edited Jan 15 at 9:22
Mostafa Ayaz
15.5k3939
asked Jan 15 at 7:58
Vibhor KumarVibhor Kumar
61
$endgroup$
5
$begingroup$
Avoid ASAP and apply logarithm in both sides wrt $e$
$endgroup$
– lab bhattacharjee
Jan 15 at 7:59
add a comment |
$begingroup$
The following set of equations is given:
$$begin{cases}{{x^2z^3over y}=e^8\{y^2zover x}=e^4\{x^2yover z^4}=1}end{cases}$$
I have solved problems before with three variables in three equations using determinant method, that is Cramer's rule. But in this problem the variables are in multiplication. I seem to have no idea on how to solve this. Please help ASAP.
matrices determinant
edited Jan 15 at 9:22
Mostafa Ayaz
15.5k3939
asked Jan 15 at 7:58
Vibhor KumarVibhor Kumar
61
$endgroup$
The following set of equations is given:
$$begin{cases}{{x^2z^3over y}=e^8\{y^2zover x}=e^4\{x^2yover z^4}=1}end{cases}$$
I have solved problems before with three variables in three equations using determinant method, that is Cramer's rule. But in this problem the variables are in multiplication. I seem to have no idea on how to solve this. Please help ASAP.
matrices determinant
matrices determinant
edited Jan 15 at 9:22
Mostafa Ayaz
15.5k3939
asked Jan 15 at 7:58
Vibhor KumarVibhor Kumar
61
edited Jan 15 at 9:22
Mostafa Ayaz
15.5k3939
asked Jan 15 at 7:58
Vibhor KumarVibhor Kumar
61
edited Jan 15 at 9:22
Mostafa Ayaz
15.5k3939
edited Jan 15 at 9:22
Mostafa Ayaz
15.5k3939
edited Jan 15 at 9:22
Mostafa Ayaz
15.5k3939
15.5k3939
asked Jan 15 at 7:58
Vibhor KumarVibhor Kumar
61
asked Jan 15 at 7:58
Vibhor KumarVibhor Kumar
61
asked Jan 15 at 7:58
Vibhor KumarVibhor Kumar
61
61
5
$begingroup$
Avoid ASAP and apply logarithm in both sides wrt $e$
$endgroup$
– lab bhattacharjee
Jan 15 at 7:59
add a comment |
5
$begingroup$
Avoid ASAP and apply logarithm in both sides wrt $e$
$endgroup$
– lab bhattacharjee
Jan 15 at 7:59
5
5
$begingroup$
Avoid ASAP and apply logarithm in both sides wrt $e$
$endgroup$
– lab bhattacharjee
Jan 15 at 7:59
$begingroup$
Avoid ASAP and apply logarithm in both sides wrt $e$
$endgroup$
– lab bhattacharjee
Jan 15 at 7:59
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Multiplying all equations together we get
$$e^{12}=x^4y^2$$ so $$y=pmfrac{e^6}{x^2}$$ plugging this in the second equation we get
$$z=frac{x^5}{e^8}$$ plugging this in the third equation we get
$$x^4=e^{-6}$$
Can you finish?
answered Jan 15 at 9:03
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.9k42865
$endgroup$
add a comment |
$begingroup$
Hint
Take $log$ from both sides of each of the equations above and define $$a=log x\b=log y\c=log z$$then you are able to use Cramer's rule again for $a,b,c$ and then find $x,y,z$.
answered Jan 15 at 9:19
Mostafa AyazMostafa Ayaz
15.5k3939
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Multiplying all equations together we get
$$e^{12}=x^4y^2$$ so $$y=pmfrac{e^6}{x^2}$$ plugging this in the second equation we get
$$z=frac{x^5}{e^8}$$ plugging this in the third equation we get
$$x^4=e^{-6}$$
Can you finish?
answered Jan 15 at 9:03
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.9k42865
$endgroup$
add a comment |
$begingroup$
Multiplying all equations together we get
$$e^{12}=x^4y^2$$ so $$y=pmfrac{e^6}{x^2}$$ plugging this in the second equation we get
$$z=frac{x^5}{e^8}$$ plugging this in the third equation we get
$$x^4=e^{-6}$$
Can you finish?
answered Jan 15 at 9:03
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.9k42865
$endgroup$
add a comment |
$begingroup$
Multiplying all equations together we get
$$e^{12}=x^4y^2$$ so $$y=pmfrac{e^6}{x^2}$$ plugging this in the second equation we get
$$z=frac{x^5}{e^8}$$ plugging this in the third equation we get
$$x^4=e^{-6}$$
Can you finish?
answered Jan 15 at 9:03
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.9k42865
$endgroup$
Multiplying all equations together we get
$$e^{12}=x^4y^2$$ so $$y=pmfrac{e^6}{x^2}$$ plugging this in the second equation we get
$$z=frac{x^5}{e^8}$$ plugging this in the third equation we get
$$x^4=e^{-6}$$
Can you finish?
answered Jan 15 at 9:03
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.9k42865
answered Jan 15 at 9:03
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.9k42865
answered Jan 15 at 9:03
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.9k42865
answered Jan 15 at 9:03
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.9k42865
74.9k42865
add a comment |
add a comment |
$begingroup$
Hint
Take $log$ from both sides of each of the equations above and define $$a=log x\b=log y\c=log z$$then you are able to use Cramer's rule again for $a,b,c$ and then find $x,y,z$.
answered Jan 15 at 9:19
Mostafa AyazMostafa Ayaz
15.5k3939
$endgroup$
add a comment |
$begingroup$
Hint
Take $log$ from both sides of each of the equations above and define $$a=log x\b=log y\c=log z$$then you are able to use Cramer's rule again for $a,b,c$ and then find $x,y,z$.
answered Jan 15 at 9:19
Mostafa AyazMostafa Ayaz
15.5k3939
$endgroup$
add a comment |
$begingroup$
Hint
Take $log$ from both sides of each of the equations above and define $$a=log x\b=log y\c=log z$$then you are able to use Cramer's rule again for $a,b,c$ and then find $x,y,z$.
answered Jan 15 at 9:19
Mostafa AyazMostafa Ayaz
15.5k3939
$endgroup$
Hint
Take $log$ from both sides of each of the equations above and define $$a=log x\b=log y\c=log z$$then you are able to use Cramer's rule again for $a,b,c$ and then find $x,y,z$.
answered Jan 15 at 9:19
Mostafa AyazMostafa Ayaz
15.5k3939
answered Jan 15 at 9:19
Mostafa AyazMostafa Ayaz
15.5k3939
answered Jan 15 at 9:19
Mostafa AyazMostafa Ayaz
15.5k3939
answered Jan 15 at 9:19
Mostafa AyazMostafa Ayaz
15.5k3939
15.5k3939
add a comment |
add a comment |
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5
$begingroup$
Avoid ASAP and apply logarithm in both sides wrt $e$
$endgroup$
– lab bhattacharjee
Jan 15 at 7:59