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
124
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
124
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
124
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
124
edited Nov 19 at 15:31
Glorfindel
3,38471730
asked Nov 19 at 15:20
M Do
124
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
124
asked Nov 19 at 15:20
M Do
124
asked Nov 19 at 15:20
M Do
124
124
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
3,609417
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
964319
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
10.5k2720
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
3,609417
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
3,609417
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
3,609417
$$0 = 0.5x^{frac{1}{2}}(3ln(x)+2)$$
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
3,609417
answered Nov 19 at 15:34
KM101
3,609417
answered Nov 19 at 15:34
KM101
3,609417
answered Nov 19 at 15:34
KM101
3,609417
3,609417
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
964319
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
964319
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
964319
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
964319
edited Nov 19 at 16:42
answered Nov 19 at 15:27
hamza boulahia
964319
answered Nov 19 at 15:27
hamza boulahia
964319
answered Nov 19 at 15:27
hamza boulahia
964319
964319
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
10.5k2720
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
10.5k2720
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
10.5k2720
$3 log x +2=0;$
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
10.5k2720
answered Nov 19 at 16:51
Peter Szilas
10.5k2720
answered Nov 19 at 16:51
Peter Szilas
10.5k2720
answered Nov 19 at 16:51
Peter Szilas
10.5k2720
10.5k2720
add a comment |
add a comment |
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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