Solving sets of 3 equations with 3 unknown with some logarithms thrown in.
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1
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I'm having some difficulty solving the following set of equations:
$6.16=xlog{(100y+z)}\
8.59=xlog{(250y+z)}\
12.72=xlog{(1000y+z)}$
The sum of the two different variables in the logarithm is throwing me off and I don't know how to solve it.
algebra-precalculus logarithms
asked Nov 23 at 12:51
chemguy
61
add a comment |
up vote
1
down vote
favorite
I'm having some difficulty solving the following set of equations:
$6.16=xlog{(100y+z)}\
8.59=xlog{(250y+z)}\
12.72=xlog{(1000y+z)}$
The sum of the two different variables in the logarithm is throwing me off and I don't know how to solve it.
algebra-precalculus logarithms
asked Nov 23 at 12:51
chemguy
61
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm having some difficulty solving the following set of equations:
$6.16=xlog{(100y+z)}\
8.59=xlog{(250y+z)}\
12.72=xlog{(1000y+z)}$
The sum of the two different variables in the logarithm is throwing me off and I don't know how to solve it.
algebra-precalculus logarithms
asked Nov 23 at 12:51
chemguy
61
I'm having some difficulty solving the following set of equations:
$6.16=xlog{(100y+z)}\
8.59=xlog{(250y+z)}\
12.72=xlog{(1000y+z)}$
The sum of the two different variables in the logarithm is throwing me off and I don't know how to solve it.
algebra-precalculus logarithms
algebra-precalculus logarithms
asked Nov 23 at 12:51
chemguy
61
asked Nov 23 at 12:51
chemguy
61
asked Nov 23 at 12:51
chemguy
61
asked Nov 23 at 12:51
chemguy
61
asked Nov 23 at 12:51
chemguy
61
61
add a comment |
add a comment |
1 Answer
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1
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I don't think there's a nice solution, but I did this:
Let $a=1/x$, $b=50y$, and $c=100y+z.$ Then the equations become
$$6.16a=log c$$
$$8.59a=log (c+3b)$$
$$12.72a =log (c+18b)$$
Exponentiate:
$$exp(6.16a) = c$$
$$exp(8.59a) = c+3b$$
$$exp(12.72a) = c+18b$$
Eliminate $c$:
$$exp(8.59a) = exp(6.16a)+3b$$
$$exp(12.72a) = exp(6.16a)+18b$$
Take $6$ times the frist equation and subtract the second to get rid of $b$:
$$6exp(8.59a) -exp(12.72a) = 5exp(6.16a)$$
Divide through by $exp(6.16a)$:
$$6exp(2.43a) - exp(6.56a) = 5.$$
Now get a numerical solution by your favorite method. Maple gives me $a=0$ (which won't work) and $a=0.3151672618.$
So $x = 1/a = 3.172918387.$
answered Nov 23 at 13:33
B. Goddard
18.2k21340
Thank you! But according to the solution x is supposed to be 7.31 and y = 0.05 and z = 1.62. Edit: nevermind, figured it out. The log was base 10 so using that I get the right answers. Thank you so much!
– chemguy
Nov 23 at 14:28
What base is your logarithm? OH, OK.
– B. Goddard
Nov 23 at 14:31
Is there any way of solving this without maple etc.? This might come up on a exam with only a simple calculator allowed.
– chemguy
Nov 24 at 14:18
There is Newton's Method (and the related "secant line method") in which you guess a solution and it gives you a better solution. Then you repeat until you have the accuracy you want.
– B. Goddard
Nov 24 at 15:26
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
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up vote
1
down vote
I don't think there's a nice solution, but I did this:
Let $a=1/x$, $b=50y$, and $c=100y+z.$ Then the equations become
$$6.16a=log c$$
$$8.59a=log (c+3b)$$
$$12.72a =log (c+18b)$$
Exponentiate:
$$exp(6.16a) = c$$
$$exp(8.59a) = c+3b$$
$$exp(12.72a) = c+18b$$
Eliminate $c$:
$$exp(8.59a) = exp(6.16a)+3b$$
$$exp(12.72a) = exp(6.16a)+18b$$
Take $6$ times the frist equation and subtract the second to get rid of $b$:
$$6exp(8.59a) -exp(12.72a) = 5exp(6.16a)$$
Divide through by $exp(6.16a)$:
$$6exp(2.43a) - exp(6.56a) = 5.$$
Now get a numerical solution by your favorite method. Maple gives me $a=0$ (which won't work) and $a=0.3151672618.$
So $x = 1/a = 3.172918387.$
answered Nov 23 at 13:33
B. Goddard
18.2k21340
Thank you! But according to the solution x is supposed to be 7.31 and y = 0.05 and z = 1.62. Edit: nevermind, figured it out. The log was base 10 so using that I get the right answers. Thank you so much!
– chemguy
Nov 23 at 14:28
What base is your logarithm? OH, OK.
– B. Goddard
Nov 23 at 14:31
Is there any way of solving this without maple etc.? This might come up on a exam with only a simple calculator allowed.
– chemguy
Nov 24 at 14:18
There is Newton's Method (and the related "secant line method") in which you guess a solution and it gives you a better solution. Then you repeat until you have the accuracy you want.
– B. Goddard
Nov 24 at 15:26
add a comment |
up vote
1
down vote
I don't think there's a nice solution, but I did this:
Let $a=1/x$, $b=50y$, and $c=100y+z.$ Then the equations become
$$6.16a=log c$$
$$8.59a=log (c+3b)$$
$$12.72a =log (c+18b)$$
Exponentiate:
$$exp(6.16a) = c$$
$$exp(8.59a) = c+3b$$
$$exp(12.72a) = c+18b$$
Eliminate $c$:
$$exp(8.59a) = exp(6.16a)+3b$$
$$exp(12.72a) = exp(6.16a)+18b$$
Take $6$ times the frist equation and subtract the second to get rid of $b$:
$$6exp(8.59a) -exp(12.72a) = 5exp(6.16a)$$
Divide through by $exp(6.16a)$:
$$6exp(2.43a) - exp(6.56a) = 5.$$
Now get a numerical solution by your favorite method. Maple gives me $a=0$ (which won't work) and $a=0.3151672618.$
So $x = 1/a = 3.172918387.$
answered Nov 23 at 13:33
B. Goddard
18.2k21340
Thank you! But according to the solution x is supposed to be 7.31 and y = 0.05 and z = 1.62. Edit: nevermind, figured it out. The log was base 10 so using that I get the right answers. Thank you so much!
– chemguy
Nov 23 at 14:28
What base is your logarithm? OH, OK.
– B. Goddard
Nov 23 at 14:31
Is there any way of solving this without maple etc.? This might come up on a exam with only a simple calculator allowed.
– chemguy
Nov 24 at 14:18
There is Newton's Method (and the related "secant line method") in which you guess a solution and it gives you a better solution. Then you repeat until you have the accuracy you want.
– B. Goddard
Nov 24 at 15:26
add a comment |
up vote
1
down vote
up vote
1
down vote
I don't think there's a nice solution, but I did this:
Let $a=1/x$, $b=50y$, and $c=100y+z.$ Then the equations become
$$6.16a=log c$$
$$8.59a=log (c+3b)$$
$$12.72a =log (c+18b)$$
Exponentiate:
$$exp(6.16a) = c$$
$$exp(8.59a) = c+3b$$
$$exp(12.72a) = c+18b$$
Eliminate $c$:
$$exp(8.59a) = exp(6.16a)+3b$$
$$exp(12.72a) = exp(6.16a)+18b$$
Take $6$ times the frist equation and subtract the second to get rid of $b$:
$$6exp(8.59a) -exp(12.72a) = 5exp(6.16a)$$
Divide through by $exp(6.16a)$:
$$6exp(2.43a) - exp(6.56a) = 5.$$
Now get a numerical solution by your favorite method. Maple gives me $a=0$ (which won't work) and $a=0.3151672618.$
So $x = 1/a = 3.172918387.$
answered Nov 23 at 13:33
B. Goddard
18.2k21340
I don't think there's a nice solution, but I did this:
Let $a=1/x$, $b=50y$, and $c=100y+z.$ Then the equations become
$$6.16a=log c$$
$$8.59a=log (c+3b)$$
$$12.72a =log (c+18b)$$
Exponentiate:
$$exp(6.16a) = c$$
$$exp(8.59a) = c+3b$$
$$exp(12.72a) = c+18b$$
Eliminate $c$:
$$exp(8.59a) = exp(6.16a)+3b$$
$$exp(12.72a) = exp(6.16a)+18b$$
Take $6$ times the frist equation and subtract the second to get rid of $b$:
$$6exp(8.59a) -exp(12.72a) = 5exp(6.16a)$$
Divide through by $exp(6.16a)$:
$$6exp(2.43a) - exp(6.56a) = 5.$$
Now get a numerical solution by your favorite method. Maple gives me $a=0$ (which won't work) and $a=0.3151672618.$
So $x = 1/a = 3.172918387.$
answered Nov 23 at 13:33
B. Goddard
18.2k21340
answered Nov 23 at 13:33
B. Goddard
18.2k21340
answered Nov 23 at 13:33
B. Goddard
18.2k21340
answered Nov 23 at 13:33
B. Goddard
18.2k21340
18.2k21340
Thank you! But according to the solution x is supposed to be 7.31 and y = 0.05 and z = 1.62. Edit: nevermind, figured it out. The log was base 10 so using that I get the right answers. Thank you so much!
– chemguy
Nov 23 at 14:28
What base is your logarithm? OH, OK.
– B. Goddard
Nov 23 at 14:31
Is there any way of solving this without maple etc.? This might come up on a exam with only a simple calculator allowed.
– chemguy
Nov 24 at 14:18
There is Newton's Method (and the related "secant line method") in which you guess a solution and it gives you a better solution. Then you repeat until you have the accuracy you want.
– B. Goddard
Nov 24 at 15:26
add a comment |
Thank you! But according to the solution x is supposed to be 7.31 and y = 0.05 and z = 1.62. Edit: nevermind, figured it out. The log was base 10 so using that I get the right answers. Thank you so much!
– chemguy
Nov 23 at 14:28
What base is your logarithm? OH, OK.
– B. Goddard
Nov 23 at 14:31
Is there any way of solving this without maple etc.? This might come up on a exam with only a simple calculator allowed.
– chemguy
Nov 24 at 14:18
There is Newton's Method (and the related "secant line method") in which you guess a solution and it gives you a better solution. Then you repeat until you have the accuracy you want.
– B. Goddard
Nov 24 at 15:26
Thank you! But according to the solution x is supposed to be 7.31 and y = 0.05 and z = 1.62. Edit: nevermind, figured it out. The log was base 10 so using that I get the right answers. Thank you so much!
– chemguy
Nov 23 at 14:28
Thank you! But according to the solution x is supposed to be 7.31 and y = 0.05 and z = 1.62. Edit: nevermind, figured it out. The log was base 10 so using that I get the right answers. Thank you so much!
– chemguy
Nov 23 at 14:28
What base is your logarithm? OH, OK.
– B. Goddard
Nov 23 at 14:31
What base is your logarithm? OH, OK.
– B. Goddard
Nov 23 at 14:31
Is there any way of solving this without maple etc.? This might come up on a exam with only a simple calculator allowed.
– chemguy
Nov 24 at 14:18
Is there any way of solving this without maple etc.? This might come up on a exam with only a simple calculator allowed.
– chemguy
Nov 24 at 14:18
There is Newton's Method (and the related "secant line method") in which you guess a solution and it gives you a better solution. Then you repeat until you have the accuracy you want.
– B. Goddard
Nov 24 at 15:26
There is Newton's Method (and the related "secant line method") in which you guess a solution and it gives you a better solution. Then you repeat until you have the accuracy you want.
– B. Goddard
Nov 24 at 15:26
add a comment |
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