Simultaneous equation with two unknowns and log

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A mathematical model for a function is $y=log_ax-b$. If $y=-21.6429$ when $x=105$ and $y=-21.1395$ when $x=211$, find $a$ and $b$ to the nearest integer

Created two equations:
$$-21.6429=log_a105-b$$
$$-21.1395=log_a211-b$$
Changed equations to:
$$a^-21.6429=105-b$$
$$a^-21.1395=211-b$$
Subtracted bottom equation from top:
$$a^-21.1395-a^-21.6429=106$$
After that I got stuck, and tried to use the calculator solver function, but it gave an out of bounds type error.

logarithms systems-of-equations

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edited Jul 22 at 2:47

Parcly Taxel

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asked Jul 22 at 2:36

user1495024

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up vote
1
down vote

favorite

A mathematical model for a function is $y=log_ax-b$. If $y=-21.6429$ when $x=105$ and $y=-21.1395$ when $x=211$, find $a$ and $b$ to the nearest integer

Created two equations:
$$-21.6429=log_a105-b$$
$$-21.1395=log_a211-b$$
Changed equations to:
$$a^-21.6429=105-b$$
$$a^-21.1395=211-b$$
Subtracted bottom equation from top:
$$a^-21.1395-a^-21.6429=106$$
After that I got stuck, and tried to use the calculator solver function, but it gave an out of bounds type error.

logarithms systems-of-equations

share|cite|improve this question

edited Jul 22 at 2:47

Parcly Taxel

33.6k136588

asked Jul 22 at 2:36

user1495024

81119

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

A mathematical model for a function is $y=log_ax-b$. If $y=-21.6429$ when $x=105$ and $y=-21.1395$ when $x=211$, find $a$ and $b$ to the nearest integer

Created two equations:
$$-21.6429=log_a105-b$$
$$-21.1395=log_a211-b$$
Changed equations to:
$$a^-21.6429=105-b$$
$$a^-21.1395=211-b$$
Subtracted bottom equation from top:
$$a^-21.1395-a^-21.6429=106$$
After that I got stuck, and tried to use the calculator solver function, but it gave an out of bounds type error.

logarithms systems-of-equations

share|cite|improve this question

edited Jul 22 at 2:47

Parcly Taxel

33.6k136588

asked Jul 22 at 2:36

user1495024

81119

A mathematical model for a function is $y=log_ax-b$. If $y=-21.6429$ when $x=105$ and $y=-21.1395$ when $x=211$, find $a$ and $b$ to the nearest integer

Created two equations:
$$-21.6429=log_a105-b$$
$$-21.1395=log_a211-b$$
Changed equations to:
$$a^-21.6429=105-b$$
$$a^-21.1395=211-b$$
Subtracted bottom equation from top:
$$a^-21.1395-a^-21.6429=106$$
After that I got stuck, and tried to use the calculator solver function, but it gave an out of bounds type error.

logarithms systems-of-equations

share|cite|improve this question

edited Jul 22 at 2:47

Parcly Taxel

33.6k136588

asked Jul 22 at 2:36

user1495024

81119

share|cite|improve this question

share|cite|improve this question

edited Jul 22 at 2:47

Parcly Taxel

33.6k136588

edited Jul 22 at 2:47

Parcly Taxel

33.6k136588

edited Jul 22 at 2:47

Parcly Taxel

33.6k136588

33.6k136588

asked Jul 22 at 2:36

user1495024

81119

asked Jul 22 at 2:36

user1495024

81119

asked Jul 22 at 2:36

user1495024

81119

81119

add a comment | 

add a comment | 

2 Answers
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You should, right from the start, subtract one equation from the other (here I'll subtract the first from the second). This cancels out $b$:
$$log_a211-log_a105=log_afrac211105=-21.1395+21.6429=0.5034$$
$$frac211105=a^0.5034$$
$$a=left(frac211105right)^1/0.5034=4.00029ldotsapprox4$$
Then we can substitute and derive $b$:
$$b=log_4105+21.6429=24.99984ldotsapprox25$$
The model is $y=log_4x-25$.

share|cite|improve this answer

answered Jul 22 at 2:57

Parcly Taxel

33.6k136588

add a comment | 

up vote
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Let $k_1=-21.6429, k_2=-21.1395,m_1=105,m_2=211$, then you have this system of equation:
$$k_1=fraclog left(m_2right)log (a)-b\
k_2=fraclog left(m_2right)log (a)-b
$$
Which gives us:
$$k_1 log a-log m_1=k_2log a-log m_2\
log a=fraclog m_1-log m_2k_1-k_2implies a=expleft(fraclog m_1-log m_2k_1-k_2right)$$
Substituting $a$ in either one of the first two equations we get:
$$beginalign
b&=fraclog m_2-k_2log alog a\
&=fraclog m_2-kleft(fraclog m_1-log m_2k_1-k_2right)fraclog m_1-log m_2k_1-k_2\
&=frac(k_1-k_2)log m_2-k_2log m_1log m_1-log m_2\
&=frack_1log m_2-k_2log m_1log m_1-log m_2
endalign$$
Substituting the actual values, you'll get:
$$ato 4.0003,bto 24.9998$$

share|cite|improve this answer

answered Jul 22 at 3:01

John Glenn

1,656223

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks John, chose other one because it was easier to understand
  – user1495024
  Jul 22 at 4:27
  

  
  
  
  

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

up vote
1
down vote



accepted

You should, right from the start, subtract one equation from the other (here I'll subtract the first from the second). This cancels out $b$:
$$log_a211-log_a105=log_afrac211105=-21.1395+21.6429=0.5034$$
$$frac211105=a^0.5034$$
$$a=left(frac211105right)^1/0.5034=4.00029ldotsapprox4$$
Then we can substitute and derive $b$:
$$b=log_4105+21.6429=24.99984ldotsapprox25$$
The model is $y=log_4x-25$.

share|cite|improve this answer

answered Jul 22 at 2:57

Parcly Taxel

33.6k136588

add a comment | 

up vote
1
down vote



accepted

You should, right from the start, subtract one equation from the other (here I'll subtract the first from the second). This cancels out $b$:
$$log_a211-log_a105=log_afrac211105=-21.1395+21.6429=0.5034$$
$$frac211105=a^0.5034$$
$$a=left(frac211105right)^1/0.5034=4.00029ldotsapprox4$$
Then we can substitute and derive $b$:
$$b=log_4105+21.6429=24.99984ldotsapprox25$$
The model is $y=log_4x-25$.

share|cite|improve this answer

answered Jul 22 at 2:57

Parcly Taxel

33.6k136588

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

You should, right from the start, subtract one equation from the other (here I'll subtract the first from the second). This cancels out $b$:
$$log_a211-log_a105=log_afrac211105=-21.1395+21.6429=0.5034$$
$$frac211105=a^0.5034$$
$$a=left(frac211105right)^1/0.5034=4.00029ldotsapprox4$$
Then we can substitute and derive $b$:
$$b=log_4105+21.6429=24.99984ldotsapprox25$$
The model is $y=log_4x-25$.

share|cite|improve this answer

answered Jul 22 at 2:57

Parcly Taxel

33.6k136588

You should, right from the start, subtract one equation from the other (here I'll subtract the first from the second). This cancels out $b$:
$$log_a211-log_a105=log_afrac211105=-21.1395+21.6429=0.5034$$
$$frac211105=a^0.5034$$
$$a=left(frac211105right)^1/0.5034=4.00029ldotsapprox4$$
Then we can substitute and derive $b$:
$$b=log_4105+21.6429=24.99984ldotsapprox25$$
The model is $y=log_4x-25$.

share|cite|improve this answer

answered Jul 22 at 2:57

Parcly Taxel

33.6k136588

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 2:57

Parcly Taxel

33.6k136588

answered Jul 22 at 2:57

Parcly Taxel

33.6k136588

answered Jul 22 at 2:57

Parcly Taxel

33.6k136588

33.6k136588

add a comment | 

add a comment | 

up vote
0
down vote

Let $k_1=-21.6429, k_2=-21.1395,m_1=105,m_2=211$, then you have this system of equation:
$$k_1=fraclog left(m_2right)log (a)-b\
k_2=fraclog left(m_2right)log (a)-b
$$
Which gives us:
$$k_1 log a-log m_1=k_2log a-log m_2\
log a=fraclog m_1-log m_2k_1-k_2implies a=expleft(fraclog m_1-log m_2k_1-k_2right)$$
Substituting $a$ in either one of the first two equations we get:
$$beginalign
b&=fraclog m_2-k_2log alog a\
&=fraclog m_2-kleft(fraclog m_1-log m_2k_1-k_2right)fraclog m_1-log m_2k_1-k_2\
&=frac(k_1-k_2)log m_2-k_2log m_1log m_1-log m_2\
&=frack_1log m_2-k_2log m_1log m_1-log m_2
endalign$$
Substituting the actual values, you'll get:
$$ato 4.0003,bto 24.9998$$

share|cite|improve this answer

answered Jul 22 at 3:01

John Glenn

1,656223

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks John, chose other one because it was easier to understand
  – user1495024
  Jul 22 at 4:27
  

  
  
  
  

add a comment | 

up vote
0
down vote

Let $k_1=-21.6429, k_2=-21.1395,m_1=105,m_2=211$, then you have this system of equation:
$$k_1=fraclog left(m_2right)log (a)-b\
k_2=fraclog left(m_2right)log (a)-b
$$
Which gives us:
$$k_1 log a-log m_1=k_2log a-log m_2\
log a=fraclog m_1-log m_2k_1-k_2implies a=expleft(fraclog m_1-log m_2k_1-k_2right)$$
Substituting $a$ in either one of the first two equations we get:
$$beginalign
b&=fraclog m_2-k_2log alog a\
&=fraclog m_2-kleft(fraclog m_1-log m_2k_1-k_2right)fraclog m_1-log m_2k_1-k_2\
&=frac(k_1-k_2)log m_2-k_2log m_1log m_1-log m_2\
&=frack_1log m_2-k_2log m_1log m_1-log m_2
endalign$$
Substituting the actual values, you'll get:
$$ato 4.0003,bto 24.9998$$

share|cite|improve this answer

answered Jul 22 at 3:01

John Glenn

1,656223

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks John, chose other one because it was easier to understand
  – user1495024
  Jul 22 at 4:27
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

Let $k_1=-21.6429, k_2=-21.1395,m_1=105,m_2=211$, then you have this system of equation:
$$k_1=fraclog left(m_2right)log (a)-b\
k_2=fraclog left(m_2right)log (a)-b
$$
Which gives us:
$$k_1 log a-log m_1=k_2log a-log m_2\
log a=fraclog m_1-log m_2k_1-k_2implies a=expleft(fraclog m_1-log m_2k_1-k_2right)$$
Substituting $a$ in either one of the first two equations we get:
$$beginalign
b&=fraclog m_2-k_2log alog a\
&=fraclog m_2-kleft(fraclog m_1-log m_2k_1-k_2right)fraclog m_1-log m_2k_1-k_2\
&=frac(k_1-k_2)log m_2-k_2log m_1log m_1-log m_2\
&=frack_1log m_2-k_2log m_1log m_1-log m_2
endalign$$
Substituting the actual values, you'll get:
$$ato 4.0003,bto 24.9998$$

share|cite|improve this answer

answered Jul 22 at 3:01

John Glenn

1,656223

Let $k_1=-21.6429, k_2=-21.1395,m_1=105,m_2=211$, then you have this system of equation:
$$k_1=fraclog left(m_2right)log (a)-b\
k_2=fraclog left(m_2right)log (a)-b
$$
Which gives us:
$$k_1 log a-log m_1=k_2log a-log m_2\
log a=fraclog m_1-log m_2k_1-k_2implies a=expleft(fraclog m_1-log m_2k_1-k_2right)$$
Substituting $a$ in either one of the first two equations we get:
$$beginalign
b&=fraclog m_2-k_2log alog a\
&=fraclog m_2-kleft(fraclog m_1-log m_2k_1-k_2right)fraclog m_1-log m_2k_1-k_2\
&=frac(k_1-k_2)log m_2-k_2log m_1log m_1-log m_2\
&=frack_1log m_2-k_2log m_1log m_1-log m_2
endalign$$
Substituting the actual values, you'll get:
$$ato 4.0003,bto 24.9998$$

share|cite|improve this answer

answered Jul 22 at 3:01

John Glenn

1,656223

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 3:01

John Glenn

1,656223

answered Jul 22 at 3:01

John Glenn

1,656223

answered Jul 22 at 3:01

John Glenn

1,656223

1,656223

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks John, chose other one because it was easier to understand
  – user1495024
  Jul 22 at 4:27
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks John, chose other one because it was easier to understand
  – user1495024
  Jul 22 at 4:27
  

  
  
  
  

Thanks John, chose other one because it was easier to understand
– user1495024
Jul 22 at 4:27

Thanks John, chose other one because it was easier to understand
– user1495024
Jul 22 at 4:27

add a comment | 

 

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