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 Nov 19 at 15:31
Glorfindel
3,38471730
asked Nov 19 at 15:20
M Do
54
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 Nov 19 at 15:31
Glorfindel
3,38471730
asked Nov 19 at 15:20
M Do
54
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
Nov 19 at 15:36
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 Nov 19 at 15:31
Glorfindel
3,38471730
asked Nov 19 at 15:20
M Do
54
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 Nov 19 at 15:31
Glorfindel
3,38471730
asked Nov 19 at 15:20
M Do
54
edited Nov 19 at 15:31
Glorfindel
3,38471730
asked Nov 19 at 15:20
M Do
54
edited Nov 19 at 15:31
Glorfindel
3,38471730
edited Nov 19 at 15:31
Glorfindel
3,38471730
edited Nov 19 at 15:31
Glorfindel
3,38471730
3,38471730
asked Nov 19 at 15:20
M Do
54
asked Nov 19 at 15:20
M Do
54
asked Nov 19 at 15:20
M Do
54
54
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
Nov 19 at 15:36
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
Nov 19 at 15:36
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
Nov 19 at 15:36
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
Nov 19 at 15:36
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 Nov 19 at 15:34
KM101
2,928416
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 Nov 19 at 15:27
hamza boulahia
954319
1
$x$ cannot be zero because $ln x$ is undefined there
– Vasya
Nov 19 at 15:31
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 Nov 19 at 16:51
Peter Szilas
10.3k2720
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 Nov 19 at 15:34
KM101
2,928416
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 Nov 19 at 15:34
KM101
2,928416
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 Nov 19 at 15:34
KM101
2,928416
$$0 = 0.5x^{frac{1}{2}}(3ln(x)+2)$$
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 Nov 19 at 15:34
KM101
2,928416
answered Nov 19 at 15:34
KM101
2,928416
answered Nov 19 at 15:34
KM101
2,928416
answered Nov 19 at 15:34
KM101
2,928416
2,928416
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 Nov 19 at 15:27
hamza boulahia
954319
1
$x$ cannot be zero because $ln x$ is undefined there
– Vasya
Nov 19 at 15:31
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 Nov 19 at 15:27
hamza boulahia
954319
1
$x$ cannot be zero because $ln x$ is undefined there
– Vasya
Nov 19 at 15:31
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 Nov 19 at 15:27
hamza boulahia
954319
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 Nov 19 at 15:27
hamza boulahia
954319
edited Nov 19 at 16:42
answered Nov 19 at 15:27
hamza boulahia
954319
answered Nov 19 at 15:27
hamza boulahia
954319
answered Nov 19 at 15:27
hamza boulahia
954319
954319
1
$x$ cannot be zero because $ln x$ is undefined there
– Vasya
Nov 19 at 15:31
add a comment |
1
$x$ cannot be zero because $ln x$ is undefined there
– Vasya
Nov 19 at 15:31
1
1
$x$ cannot be zero because $ln x$ is undefined there
– Vasya
Nov 19 at 15:31
$x$ cannot be zero because $ln x$ is undefined there
– Vasya
Nov 19 at 15:31
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 Nov 19 at 16:51
Peter Szilas
10.3k2720
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 Nov 19 at 16:51
Peter Szilas
10.3k2720
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 Nov 19 at 16:51
Peter Szilas
10.3k2720
$3 log x +2=0;$
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 Nov 19 at 16:51
Peter Szilas
10.3k2720
answered Nov 19 at 16:51
Peter Szilas
10.3k2720
answered Nov 19 at 16:51
Peter Szilas
10.3k2720
answered Nov 19 at 16:51
Peter Szilas
10.3k2720
10.3k2720
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
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
Nov 19 at 15:36