Mixing
Trouble factoring with $ln$
up vote
0
down vote
favorite
I have the equation
$$0 = 0.5x^frac {1}{2}(3ln(x) + 2) $$
How do I interpret the root inside the brackets?
The solutions are $x = 0$ and $x = e^{-frac{2}{3}}$, but I have absolutely no idea how that last one was found. Could anybody explain it to me?
logarithms factoring
edited yesterday
Glorfindel
3,38471730
asked yesterday
M Do
91
add a comment |
up vote
0
down vote
favorite
I have the equation
$$0 = 0.5x^frac {1}{2}(3ln(x) + 2) $$
How do I interpret the root inside the brackets?
The solutions are $x = 0$ and $x = e^{-frac{2}{3}}$, but I have absolutely no idea how that last one was found. Could anybody explain it to me?
logarithms factoring
edited yesterday
Glorfindel
3,38471730
asked yesterday
M Do
91
1
Recall the definition of $ln x$. $$ln x = y iff e^y = x$$ Also, the first solution is incorrect since $x = 0$ is not within the domain $x in (0, +infty)$ in which $ln x$ is defined.
– KM101
yesterday
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have the equation
$$0 = 0.5x^frac {1}{2}(3ln(x) + 2) $$
How do I interpret the root inside the brackets?
The solutions are $x = 0$ and $x = e^{-frac{2}{3}}$, but I have absolutely no idea how that last one was found. Could anybody explain it to me?
logarithms factoring
edited yesterday
Glorfindel
3,38471730
asked yesterday
M Do
91
I have the equation
$$0 = 0.5x^frac {1}{2}(3ln(x) + 2) $$
How do I interpret the root inside the brackets?
The solutions are $x = 0$ and $x = e^{-frac{2}{3}}$, but I have absolutely no idea how that last one was found. Could anybody explain it to me?
logarithms factoring
logarithms factoring
edited yesterday
Glorfindel
3,38471730
asked yesterday
M Do
91
edited yesterday
Glorfindel
3,38471730
asked yesterday
M Do
91
edited yesterday
Glorfindel
3,38471730
edited yesterday
Glorfindel
3,38471730
edited yesterday
Glorfindel
3,38471730
3,38471730
asked yesterday
M Do
91
asked yesterday
M Do
91
asked yesterday
M Do
91
91
1
Recall the definition of $ln x$. $$ln x = y iff e^y = x$$ Also, the first solution is incorrect since $x = 0$ is not within the domain $x in (0, +infty)$ in which $ln x$ is defined.
– KM101
yesterday
add a comment |
1
Recall the definition of $ln x$. $$ln x = y iff e^y = x$$ Also, the first solution is incorrect since $x = 0$ is not within the domain $x in (0, +infty)$ in which $ln x$ is defined.
– KM101
yesterday
1
1
Recall the definition of $ln x$. $$ln x = y iff e^y = x$$ Also, the first solution is incorrect since $x = 0$ is not within the domain $x in (0, +infty)$ in which $ln x$ is defined.
– KM101
yesterday
Recall the definition of $ln x$. $$ln x = y iff e^y = x$$ Also, the first solution is incorrect since $x = 0$ is not within the domain $x in (0, +infty)$ in which $ln x$ is defined.
– KM101
yesterday
add a comment |
3 Answers
3
active
oldest
votes
up vote
1
down vote
$$0 = 0.5x^{frac{1}{2}}(3ln(x)+2)$$
Set either factor equal to $0$. So, you get
$$0.5x^{frac{1}{2}} = 0 implies x = 0$$
or
$$3ln(x)+2 = 0 implies 3ln x = -2 implies ln x = -frac{2}{3} implies x = e^{-frac{2}{3}}$$
However, since $ln x$ is defined for all $x > 0$, the first solution is discarded, leaving you with only one solution.
answered yesterday
KM101
1,823313
add a comment |
up vote
0
down vote
We have, $$ 0 = 0.5x^frac {1}{2}(3ln(x) + 2)$$
Then,
$$ 0.5x^frac {1}{2} quad mbox{or}quad 3ln(x) + 2=0$$
hence,
$$ x=0quad mbox{or}quad ln(x)=frac{-2}{3}Longrightarrow x=e^frac{-2}{3}$$
And since $ln(x)$ is only defined on $Bbb{R}^*_+$ then the only solution is:
$$ x=e^frac{-2}{3}$$
answered yesterday
hamza boulahia
931319
1
$x$ cannot be zero because $ln x$ is undefined there
– Vasya
yesterday
add a comment |
up vote
0
down vote
$3 log x +2=0;$
Since $x >0$ (why?) , we set
$x=e^y$ , $y>0$, real.
Then:
$3 log (e^y) +2=0;$
$3y +2=0;$
$y=-2/3.$
Since $x=e^y$, we get
$x=e^{-2/3}.$
answered yesterday
Peter Szilas
9,9292720
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
$$0 = 0.5x^{frac{1}{2}}(3ln(x)+2)$$
Set either factor equal to $0$. So, you get
$$0.5x^{frac{1}{2}} = 0 implies x = 0$$
or
$$3ln(x)+2 = 0 implies 3ln x = -2 implies ln x = -frac{2}{3} implies x = e^{-frac{2}{3}}$$
However, since $ln x$ is defined for all $x > 0$, the first solution is discarded, leaving you with only one solution.
answered yesterday
KM101
1,823313
add a comment |
up vote
1
down vote
$$0 = 0.5x^{frac{1}{2}}(3ln(x)+2)$$
Set either factor equal to $0$. So, you get
$$0.5x^{frac{1}{2}} = 0 implies x = 0$$
or
$$3ln(x)+2 = 0 implies 3ln x = -2 implies ln x = -frac{2}{3} implies x = e^{-frac{2}{3}}$$
However, since $ln x$ is defined for all $x > 0$, the first solution is discarded, leaving you with only one solution.
answered yesterday
KM101
1,823313
add a comment |
up vote
1
down vote
up vote
1
down vote
$$0 = 0.5x^{frac{1}{2}}(3ln(x)+2)$$
Set either factor equal to $0$. So, you get
$$0.5x^{frac{1}{2}} = 0 implies x = 0$$
or
$$3ln(x)+2 = 0 implies 3ln x = -2 implies ln x = -frac{2}{3} implies x = e^{-frac{2}{3}}$$
However, since $ln x$ is defined for all $x > 0$, the first solution is discarded, leaving you with only one solution.
answered yesterday
KM101
1,823313
$$0 = 0.5x^{frac{1}{2}}(3ln(x)+2)$$
Set either factor equal to $0$. So, you get
$$0.5x^{frac{1}{2}} = 0 implies x = 0$$
or
$$3ln(x)+2 = 0 implies 3ln x = -2 implies ln x = -frac{2}{3} implies x = e^{-frac{2}{3}}$$
However, since $ln x$ is defined for all $x > 0$, the first solution is discarded, leaving you with only one solution.
answered yesterday
KM101
1,823313
answered yesterday
KM101
1,823313
answered yesterday
KM101
1,823313
answered yesterday
KM101
1,823313
1,823313
add a comment |
add a comment |
up vote
0
down vote
We have, $$ 0 = 0.5x^frac {1}{2}(3ln(x) + 2)$$
Then,
$$ 0.5x^frac {1}{2} quad mbox{or}quad 3ln(x) + 2=0$$
hence,
$$ x=0quad mbox{or}quad ln(x)=frac{-2}{3}Longrightarrow x=e^frac{-2}{3}$$
And since $ln(x)$ is only defined on $Bbb{R}^*_+$ then the only solution is:
$$ x=e^frac{-2}{3}$$
answered yesterday
hamza boulahia
931319
1
$x$ cannot be zero because $ln x$ is undefined there
– Vasya
yesterday
add a comment |
up vote
0
down vote
We have, $$ 0 = 0.5x^frac {1}{2}(3ln(x) + 2)$$
Then,
$$ 0.5x^frac {1}{2} quad mbox{or}quad 3ln(x) + 2=0$$
hence,
$$ x=0quad mbox{or}quad ln(x)=frac{-2}{3}Longrightarrow x=e^frac{-2}{3}$$
And since $ln(x)$ is only defined on $Bbb{R}^*_+$ then the only solution is:
$$ x=e^frac{-2}{3}$$
answered yesterday
hamza boulahia
931319
1
$x$ cannot be zero because $ln x$ is undefined there
– Vasya
yesterday
add a comment |
up vote
0
down vote
up vote
0
down vote
We have, $$ 0 = 0.5x^frac {1}{2}(3ln(x) + 2)$$
Then,
$$ 0.5x^frac {1}{2} quad mbox{or}quad 3ln(x) + 2=0$$
hence,
$$ x=0quad mbox{or}quad ln(x)=frac{-2}{3}Longrightarrow x=e^frac{-2}{3}$$
And since $ln(x)$ is only defined on $Bbb{R}^*_+$ then the only solution is:
$$ x=e^frac{-2}{3}$$
answered yesterday
hamza boulahia
931319
We have, $$ 0 = 0.5x^frac {1}{2}(3ln(x) + 2)$$
Then,
$$ 0.5x^frac {1}{2} quad mbox{or}quad 3ln(x) + 2=0$$
hence,
$$ x=0quad mbox{or}quad ln(x)=frac{-2}{3}Longrightarrow x=e^frac{-2}{3}$$
And since $ln(x)$ is only defined on $Bbb{R}^*_+$ then the only solution is:
$$ x=e^frac{-2}{3}$$
answered yesterday
hamza boulahia
931319
edited yesterday
answered yesterday
hamza boulahia
931319
answered yesterday
hamza boulahia
931319
answered yesterday
hamza boulahia
931319
931319
1
$x$ cannot be zero because $ln x$ is undefined there
– Vasya
yesterday
add a comment |
1
$x$ cannot be zero because $ln x$ is undefined there
– Vasya
yesterday
1
1
$x$ cannot be zero because $ln x$ is undefined there
– Vasya
yesterday
$x$ cannot be zero because $ln x$ is undefined there
– Vasya
yesterday
add a comment |
up vote
0
down vote
$3 log x +2=0;$
Since $x >0$ (why?) , we set
$x=e^y$ , $y>0$, real.
Then:
$3 log (e^y) +2=0;$
$3y +2=0;$
$y=-2/3.$
Since $x=e^y$, we get
$x=e^{-2/3}.$
answered yesterday
Peter Szilas
9,9292720
add a comment |
up vote
0
down vote
$3 log x +2=0;$
Since $x >0$ (why?) , we set
$x=e^y$ , $y>0$, real.
Then:
$3 log (e^y) +2=0;$
$3y +2=0;$
$y=-2/3.$
Since $x=e^y$, we get
$x=e^{-2/3}.$
answered yesterday
Peter Szilas
9,9292720
add a comment |
up vote
0
down vote
up vote
0
down vote
$3 log x +2=0;$
Since $x >0$ (why?) , we set
$x=e^y$ , $y>0$, real.
Then:
$3 log (e^y) +2=0;$
$3y +2=0;$
$y=-2/3.$
Since $x=e^y$, we get
$x=e^{-2/3}.$
answered yesterday
Peter Szilas
9,9292720
$3 log x +2=0;$
Since $x >0$ (why?) , we set
$x=e^y$ , $y>0$, real.
Then:
$3 log (e^y) +2=0;$
$3y +2=0;$
$y=-2/3.$
Since $x=e^y$, we get
$x=e^{-2/3}.$
answered yesterday
Peter Szilas
9,9292720
answered yesterday
Peter Szilas
9,9292720
answered yesterday
Peter Szilas
9,9292720
answered yesterday
Peter Szilas
9,9292720
9,9292720
add a comment |
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3005059%2ftrouble-factoring-with-ln%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
Recall the definition of $ln x$. $$ln x = y iff e^y = x$$ Also, the first solution is incorrect since $x = 0$ is not within the domain $x in (0, +infty)$ in which $ln x$ is defined.
– KM101
yesterday