Isolate $y$ (that is nested inside $ln$)
I am trying to isolate y in this equation:
$$-4/3·ln(|y-60|)=x+c$$
If I use a cas-tool to isolate $y$, I get:
$$60.-(2.71828182846)^{−0.75*x-0.75*c}=y$$
If I try isolating $y$ by hand I get:
enter image description here
These two are not the same, is the cas-tool right or am I right? What are the rules to isolate something when the absolute value is taken of it as in this case.
Proof they are not equal: (black is my result, red is cas-tool's result)
enter image description here
algebra-precalculus
asked Nov 26 at 9:15
Ryan Cameron
397
add a comment |
I am trying to isolate y in this equation:
$$-4/3·ln(|y-60|)=x+c$$
If I use a cas-tool to isolate $y$, I get:
$$60.-(2.71828182846)^{−0.75*x-0.75*c}=y$$
If I try isolating $y$ by hand I get:
enter image description here
These two are not the same, is the cas-tool right or am I right? What are the rules to isolate something when the absolute value is taken of it as in this case.
Proof they are not equal: (black is my result, red is cas-tool's result)
enter image description here
algebra-precalculus
asked Nov 26 at 9:15
Ryan Cameron
397
Domain of $ln$ is $(0,infty)$.
– Yadati Kiran
Nov 26 at 9:24
@YadatiKiran What is your point? What is the correct result?
– Ryan Cameron
Nov 26 at 12:06
You have two cases; one where $y-60 = ldots$ and $-(y-60)= ldots$. Since htere is no requirement on $y$, consider these two cases in the context of the range of $x$ and the domain of $y$.
– Kevin
Nov 26 at 12:26
add a comment |
I am trying to isolate y in this equation:
$$-4/3·ln(|y-60|)=x+c$$
If I use a cas-tool to isolate $y$, I get:
$$60.-(2.71828182846)^{−0.75*x-0.75*c}=y$$
If I try isolating $y$ by hand I get:
enter image description here
These two are not the same, is the cas-tool right or am I right? What are the rules to isolate something when the absolute value is taken of it as in this case.
Proof they are not equal: (black is my result, red is cas-tool's result)
enter image description here
algebra-precalculus
asked Nov 26 at 9:15
Ryan Cameron
397
I am trying to isolate y in this equation:
$$-4/3·ln(|y-60|)=x+c$$
If I use a cas-tool to isolate $y$, I get:
$$60.-(2.71828182846)^{−0.75*x-0.75*c}=y$$
If I try isolating $y$ by hand I get:
enter image description here
These two are not the same, is the cas-tool right or am I right? What are the rules to isolate something when the absolute value is taken of it as in this case.
Proof they are not equal: (black is my result, red is cas-tool's result)
enter image description here
algebra-precalculus
algebra-precalculus
asked Nov 26 at 9:15
Ryan Cameron
397
asked Nov 26 at 9:15
Ryan Cameron
397
edited Nov 26 at 12:08
asked Nov 26 at 9:15
Ryan Cameron
397
asked Nov 26 at 9:15
Ryan Cameron
397
asked Nov 26 at 9:15
Ryan Cameron
397
397
Domain of $ln$ is $(0,infty)$.
– Yadati Kiran
Nov 26 at 9:24
@YadatiKiran What is your point? What is the correct result?
– Ryan Cameron
Nov 26 at 12:06
You have two cases; one where $y-60 = ldots$ and $-(y-60)= ldots$. Since htere is no requirement on $y$, consider these two cases in the context of the range of $x$ and the domain of $y$.
– Kevin
Nov 26 at 12:26
add a comment |
Domain of $ln$ is $(0,infty)$.
– Yadati Kiran
Nov 26 at 9:24
@YadatiKiran What is your point? What is the correct result?
– Ryan Cameron
Nov 26 at 12:06
You have two cases; one where $y-60 = ldots$ and $-(y-60)= ldots$. Since htere is no requirement on $y$, consider these two cases in the context of the range of $x$ and the domain of $y$.
– Kevin
Nov 26 at 12:26
Domain of $ln$ is $(0,infty)$.
– Yadati Kiran
Nov 26 at 9:24
Domain of $ln$ is $(0,infty)$.
– Yadati Kiran
Nov 26 at 9:24
@YadatiKiran What is your point? What is the correct result?
– Ryan Cameron
Nov 26 at 12:06
@YadatiKiran What is your point? What is the correct result?
– Ryan Cameron
Nov 26 at 12:06
You have two cases; one where $y-60 = ldots$ and $-(y-60)= ldots$. Since htere is no requirement on $y$, consider these two cases in the context of the range of $x$ and the domain of $y$.
– Kevin
Nov 26 at 12:26
You have two cases; one where $y-60 = ldots$ and $-(y-60)= ldots$. Since htere is no requirement on $y$, consider these two cases in the context of the range of $x$ and the domain of $y$.
– Kevin
Nov 26 at 12:26
add a comment |
2 Answers
2
active
oldest
votes
If $y>60$, then these are equivalent:
$$
-4/3·ln(|y-60|)=x+c \
-4/3·ln(y-60)=x+c \
ln(y-60)=frac{-3(x+c)}{4} \
y-60=expfrac{-3(x+c)}{4} \
y=60+expfrac{-3(x+c)}{4}\
$$
Similarly, if $y<60$,
$$
-4/3·ln(|y-60|)=x+c \
-4/3·ln(60-y)=x+c \
ln(60-y)=frac{-3(x+c)}{4} \
60-y=expfrac{-3(x+c)}{4} \
y=60-expfrac{-3(x+c)}{4}\
$$
Note the $2.71828$ in the cas answer is $e$, $0.75 = frac{3}{4}$ and $exp(z) = e^z$. So the caz answer is my second solution.
answered Nov 26 at 12:24
GEdgar
61.6k267168
add a comment |
First, multiply everything by $frac{-3}{4}$ to obtain
$$ln|y-60|=frac{-3(x+c)}{4}$$
Now exponentiate everything:
$$|y-60|$ = e^{frac{-3(x+c)}{4}}$$
Now, we need to take cases, because of the absolute value sign (which is what you lost along the way in your derivation):
If $y > 60$, then
$$y = 60 + e^{frac{-3(x+c)}{4}}$$
which agrees exactly with your answer, because the absolute value sign that you lost doesn't matter in this case.
But if $y < 60$, then
$$y = 60 - e^{frac{-3(x+c)}{4}}$$
which agrees with the value given by your CAS tool (up to the CAS tool's rounding). I assume that the CAS tool is assuming some default values of $y$ to address the absolute value sign. Probably $y = 0$ or something.
If you feel like putting everything into one expression, then, it's
$$y = 60+mathrm{mathop{sgn}}(y-60)e^frac{-3(x+c)}{4}.$$ (or something equivalent, where $mathrm{sgn}$ is the sign function, taking the value $1$ when the argument is greater than $0$, and the value $-1$ when its argument is less than $0$: the value when the argument is exactly $0$ is usually taken to be $0$, but it doesn't matter here, because your expression isn't defined when $y = 60$ anyway.
answered Nov 26 at 12:25
user3482749
2,381414
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
If $y>60$, then these are equivalent:
$$
-4/3·ln(|y-60|)=x+c \
-4/3·ln(y-60)=x+c \
ln(y-60)=frac{-3(x+c)}{4} \
y-60=expfrac{-3(x+c)}{4} \
y=60+expfrac{-3(x+c)}{4}\
$$
Similarly, if $y<60$,
$$
-4/3·ln(|y-60|)=x+c \
-4/3·ln(60-y)=x+c \
ln(60-y)=frac{-3(x+c)}{4} \
60-y=expfrac{-3(x+c)}{4} \
y=60-expfrac{-3(x+c)}{4}\
$$
Note the $2.71828$ in the cas answer is $e$, $0.75 = frac{3}{4}$ and $exp(z) = e^z$. So the caz answer is my second solution.
answered Nov 26 at 12:24
GEdgar
61.6k267168
add a comment |
If $y>60$, then these are equivalent:
$$
-4/3·ln(|y-60|)=x+c \
-4/3·ln(y-60)=x+c \
ln(y-60)=frac{-3(x+c)}{4} \
y-60=expfrac{-3(x+c)}{4} \
y=60+expfrac{-3(x+c)}{4}\
$$
Similarly, if $y<60$,
$$
-4/3·ln(|y-60|)=x+c \
-4/3·ln(60-y)=x+c \
ln(60-y)=frac{-3(x+c)}{4} \
60-y=expfrac{-3(x+c)}{4} \
y=60-expfrac{-3(x+c)}{4}\
$$
Note the $2.71828$ in the cas answer is $e$, $0.75 = frac{3}{4}$ and $exp(z) = e^z$. So the caz answer is my second solution.
answered Nov 26 at 12:24
GEdgar
61.6k267168
add a comment |
If $y>60$, then these are equivalent:
$$
-4/3·ln(|y-60|)=x+c \
-4/3·ln(y-60)=x+c \
ln(y-60)=frac{-3(x+c)}{4} \
y-60=expfrac{-3(x+c)}{4} \
y=60+expfrac{-3(x+c)}{4}\
$$
Similarly, if $y<60$,
$$
-4/3·ln(|y-60|)=x+c \
-4/3·ln(60-y)=x+c \
ln(60-y)=frac{-3(x+c)}{4} \
60-y=expfrac{-3(x+c)}{4} \
y=60-expfrac{-3(x+c)}{4}\
$$
Note the $2.71828$ in the cas answer is $e$, $0.75 = frac{3}{4}$ and $exp(z) = e^z$. So the caz answer is my second solution.
answered Nov 26 at 12:24
GEdgar
61.6k267168
If $y>60$, then these are equivalent:
$$
-4/3·ln(|y-60|)=x+c \
-4/3·ln(y-60)=x+c \
ln(y-60)=frac{-3(x+c)}{4} \
y-60=expfrac{-3(x+c)}{4} \
y=60+expfrac{-3(x+c)}{4}\
$$
Similarly, if $y<60$,
$$
-4/3·ln(|y-60|)=x+c \
-4/3·ln(60-y)=x+c \
ln(60-y)=frac{-3(x+c)}{4} \
60-y=expfrac{-3(x+c)}{4} \
y=60-expfrac{-3(x+c)}{4}\
$$
Note the $2.71828$ in the cas answer is $e$, $0.75 = frac{3}{4}$ and $exp(z) = e^z$. So the caz answer is my second solution.
answered Nov 26 at 12:24
GEdgar
61.6k267168
answered Nov 26 at 12:24
GEdgar
61.6k267168
answered Nov 26 at 12:24
GEdgar
61.6k267168
answered Nov 26 at 12:24
GEdgar
61.6k267168
61.6k267168
add a comment |
add a comment |
First, multiply everything by $frac{-3}{4}$ to obtain
$$ln|y-60|=frac{-3(x+c)}{4}$$
Now exponentiate everything:
$$|y-60|$ = e^{frac{-3(x+c)}{4}}$$
Now, we need to take cases, because of the absolute value sign (which is what you lost along the way in your derivation):
If $y > 60$, then
$$y = 60 + e^{frac{-3(x+c)}{4}}$$
which agrees exactly with your answer, because the absolute value sign that you lost doesn't matter in this case.
But if $y < 60$, then
$$y = 60 - e^{frac{-3(x+c)}{4}}$$
which agrees with the value given by your CAS tool (up to the CAS tool's rounding). I assume that the CAS tool is assuming some default values of $y$ to address the absolute value sign. Probably $y = 0$ or something.
If you feel like putting everything into one expression, then, it's
$$y = 60+mathrm{mathop{sgn}}(y-60)e^frac{-3(x+c)}{4}.$$ (or something equivalent, where $mathrm{sgn}$ is the sign function, taking the value $1$ when the argument is greater than $0$, and the value $-1$ when its argument is less than $0$: the value when the argument is exactly $0$ is usually taken to be $0$, but it doesn't matter here, because your expression isn't defined when $y = 60$ anyway.
answered Nov 26 at 12:25
user3482749
2,381414
add a comment |
First, multiply everything by $frac{-3}{4}$ to obtain
$$ln|y-60|=frac{-3(x+c)}{4}$$
Now exponentiate everything:
$$|y-60|$ = e^{frac{-3(x+c)}{4}}$$
Now, we need to take cases, because of the absolute value sign (which is what you lost along the way in your derivation):
If $y > 60$, then
$$y = 60 + e^{frac{-3(x+c)}{4}}$$
which agrees exactly with your answer, because the absolute value sign that you lost doesn't matter in this case.
But if $y < 60$, then
$$y = 60 - e^{frac{-3(x+c)}{4}}$$
which agrees with the value given by your CAS tool (up to the CAS tool's rounding). I assume that the CAS tool is assuming some default values of $y$ to address the absolute value sign. Probably $y = 0$ or something.
If you feel like putting everything into one expression, then, it's
$$y = 60+mathrm{mathop{sgn}}(y-60)e^frac{-3(x+c)}{4}.$$ (or something equivalent, where $mathrm{sgn}$ is the sign function, taking the value $1$ when the argument is greater than $0$, and the value $-1$ when its argument is less than $0$: the value when the argument is exactly $0$ is usually taken to be $0$, but it doesn't matter here, because your expression isn't defined when $y = 60$ anyway.
answered Nov 26 at 12:25
user3482749
2,381414
add a comment |
First, multiply everything by $frac{-3}{4}$ to obtain
$$ln|y-60|=frac{-3(x+c)}{4}$$
Now exponentiate everything:
$$|y-60|$ = e^{frac{-3(x+c)}{4}}$$
Now, we need to take cases, because of the absolute value sign (which is what you lost along the way in your derivation):
If $y > 60$, then
$$y = 60 + e^{frac{-3(x+c)}{4}}$$
which agrees exactly with your answer, because the absolute value sign that you lost doesn't matter in this case.
But if $y < 60$, then
$$y = 60 - e^{frac{-3(x+c)}{4}}$$
which agrees with the value given by your CAS tool (up to the CAS tool's rounding). I assume that the CAS tool is assuming some default values of $y$ to address the absolute value sign. Probably $y = 0$ or something.
If you feel like putting everything into one expression, then, it's
$$y = 60+mathrm{mathop{sgn}}(y-60)e^frac{-3(x+c)}{4}.$$ (or something equivalent, where $mathrm{sgn}$ is the sign function, taking the value $1$ when the argument is greater than $0$, and the value $-1$ when its argument is less than $0$: the value when the argument is exactly $0$ is usually taken to be $0$, but it doesn't matter here, because your expression isn't defined when $y = 60$ anyway.
answered Nov 26 at 12:25
user3482749
2,381414
First, multiply everything by $frac{-3}{4}$ to obtain
$$ln|y-60|=frac{-3(x+c)}{4}$$
Now exponentiate everything:
$$|y-60|$ = e^{frac{-3(x+c)}{4}}$$
Now, we need to take cases, because of the absolute value sign (which is what you lost along the way in your derivation):
If $y > 60$, then
$$y = 60 + e^{frac{-3(x+c)}{4}}$$
which agrees exactly with your answer, because the absolute value sign that you lost doesn't matter in this case.
But if $y < 60$, then
$$y = 60 - e^{frac{-3(x+c)}{4}}$$
which agrees with the value given by your CAS tool (up to the CAS tool's rounding). I assume that the CAS tool is assuming some default values of $y$ to address the absolute value sign. Probably $y = 0$ or something.
If you feel like putting everything into one expression, then, it's
$$y = 60+mathrm{mathop{sgn}}(y-60)e^frac{-3(x+c)}{4}.$$ (or something equivalent, where $mathrm{sgn}$ is the sign function, taking the value $1$ when the argument is greater than $0$, and the value $-1$ when its argument is less than $0$: the value when the argument is exactly $0$ is usually taken to be $0$, but it doesn't matter here, because your expression isn't defined when $y = 60$ anyway.
answered Nov 26 at 12:25
user3482749
2,381414
answered Nov 26 at 12:25
user3482749
2,381414
answered Nov 26 at 12:25
user3482749
2,381414
answered Nov 26 at 12:25
user3482749
2,381414
2,381414
add a comment |
add a comment |
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Domain of $ln$ is $(0,infty)$.
– Yadati Kiran
Nov 26 at 9:24
@YadatiKiran What is your point? What is the correct result?
– Ryan Cameron
Nov 26 at 12:06
You have two cases; one where $y-60 = ldots$ and $-(y-60)= ldots$. Since htere is no requirement on $y$, consider these two cases in the context of the range of $x$ and the domain of $y$.
– Kevin
Nov 26 at 12:26