Mixing
How to find the unknown values of a natural log function.
$begingroup$
I was able to do (a) and (b), but couldn't find the answer for (c). I thought I would be able to find them by creating a simultaneous equation, and one of the equations I found is 4a+b=1, by using the graph. Can someone find what the other equation would be?
functions
edited Jan 17 at 15:12
N. F. Taussig
44.4k93357
asked Jan 17 at 14:10
Hami the PenguinHami the Penguin
474
$endgroup$
add a comment |
$begingroup$
I was able to do (a) and (b), but couldn't find the answer for (c). I thought I would be able to find them by creating a simultaneous equation, and one of the equations I found is 4a+b=1, by using the graph. Can someone find what the other equation would be?
functions
edited Jan 17 at 15:12
N. F. Taussig
44.4k93357
asked Jan 17 at 14:10
Hami the PenguinHami the Penguin
474
$endgroup$
add a comment |
$begingroup$
I was able to do (a) and (b), but couldn't find the answer for (c). I thought I would be able to find them by creating a simultaneous equation, and one of the equations I found is 4a+b=1, by using the graph. Can someone find what the other equation would be?
functions
edited Jan 17 at 15:12
N. F. Taussig
44.4k93357
asked Jan 17 at 14:10
Hami the PenguinHami the Penguin
474
$endgroup$
I was able to do (a) and (b), but couldn't find the answer for (c). I thought I would be able to find them by creating a simultaneous equation, and one of the equations I found is 4a+b=1, by using the graph. Can someone find what the other equation would be?
functions
functions
edited Jan 17 at 15:12
N. F. Taussig
44.4k93357
asked Jan 17 at 14:10
Hami the PenguinHami the Penguin
474
edited Jan 17 at 15:12
N. F. Taussig
44.4k93357
asked Jan 17 at 14:10
Hami the PenguinHami the Penguin
474
edited Jan 17 at 15:12
N. F. Taussig
44.4k93357
edited Jan 17 at 15:12
N. F. Taussig
44.4k93357
edited Jan 17 at 15:12
N. F. Taussig
44.4k93357
44.4k93357
asked Jan 17 at 14:10
Hami the PenguinHami the Penguin
474
asked Jan 17 at 14:10
Hami the PenguinHami the Penguin
474
asked Jan 17 at 14:10
Hami the PenguinHami the Penguin
474
474
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
For the other equation, note that the singularity -- the point where the function goes to negative infinity -- is at $x = 1$. The $log$ function goes to negative infinity when its argument goes to zero. Thus, $a times 1 + b = 0$ is the other equation.
answered Jan 17 at 14:14
Mees de VriesMees de Vries
17.3k12957
$endgroup$
$begingroup$
Thank you very much.
$endgroup$
– Hami the Penguin
Jan 17 at 14:22
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For the other equation, note that the singularity -- the point where the function goes to negative infinity -- is at $x = 1$. The $log$ function goes to negative infinity when its argument goes to zero. Thus, $a times 1 + b = 0$ is the other equation.
answered Jan 17 at 14:14
Mees de VriesMees de Vries
17.3k12957
$endgroup$
$begingroup$
Thank you very much.
$endgroup$
– Hami the Penguin
Jan 17 at 14:22
add a comment |
$begingroup$
For the other equation, note that the singularity -- the point where the function goes to negative infinity -- is at $x = 1$. The $log$ function goes to negative infinity when its argument goes to zero. Thus, $a times 1 + b = 0$ is the other equation.
answered Jan 17 at 14:14
Mees de VriesMees de Vries
17.3k12957
$endgroup$
$begingroup$
Thank you very much.
$endgroup$
– Hami the Penguin
Jan 17 at 14:22
add a comment |
$begingroup$
For the other equation, note that the singularity -- the point where the function goes to negative infinity -- is at $x = 1$. The $log$ function goes to negative infinity when its argument goes to zero. Thus, $a times 1 + b = 0$ is the other equation.
answered Jan 17 at 14:14
Mees de VriesMees de Vries
17.3k12957
$endgroup$
For the other equation, note that the singularity -- the point where the function goes to negative infinity -- is at $x = 1$. The $log$ function goes to negative infinity when its argument goes to zero. Thus, $a times 1 + b = 0$ is the other equation.
answered Jan 17 at 14:14
Mees de VriesMees de Vries
17.3k12957
answered Jan 17 at 14:14
Mees de VriesMees de Vries
17.3k12957
answered Jan 17 at 14:14
Mees de VriesMees de Vries
17.3k12957
answered Jan 17 at 14:14
Mees de VriesMees de Vries
17.3k12957
17.3k12957
$begingroup$
Thank you very much.
$endgroup$
– Hami the Penguin
Jan 17 at 14:22
add a comment |
$begingroup$
Thank you very much.
$endgroup$
– Hami the Penguin
Jan 17 at 14:22
$begingroup$
Thank you very much.
$endgroup$
– Hami the Penguin
Jan 17 at 14:22
$begingroup$
Thank you very much.
$endgroup$
– Hami the Penguin
Jan 17 at 14:22
add a comment |
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