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Domain and range of a logarithmic absolute value function.
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up vote
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down vote
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$$y = ln (|ln x|)$$
Attempt at solution:
For the function to be defined, $|ln x|> 0$
$implies ln x > 0 $
$implies x > 1$ [$because$ anti-logging both sides.]
But clearly, the function is valid if $xin(0,1)$ as well.
I am not sure what I am doing wrong. I might be doing something completely illegal here. So sorry about that!
I am looking for a mechanical way of solving these types of problems. So solutions which don't involve graphs (if possible) are requested.
Thanks in advance!
Ps. I have no clue on how to begin calculating for the range.
logarithms graphing-functions absolute-value
asked Jul 28 at 11:11
Amaimon43
11
add a comment |Â
up vote
-1
down vote
favorite
$$y = ln (|ln x|)$$
Attempt at solution:
For the function to be defined, $|ln x|> 0$
$implies ln x > 0 $
$implies x > 1$ [$because$ anti-logging both sides.]
But clearly, the function is valid if $xin(0,1)$ as well.
I am not sure what I am doing wrong. I might be doing something completely illegal here. So sorry about that!
I am looking for a mechanical way of solving these types of problems. So solutions which don't involve graphs (if possible) are requested.
Thanks in advance!
Ps. I have no clue on how to begin calculating for the range.
logarithms graphing-functions absolute-value
asked Jul 28 at 11:11
Amaimon43
11
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 28 at 11:25
With $mod$ do you mean the absolute value?
– Davide Morgante
Jul 28 at 11:39
Ya that's what I meant.
– Amaimon43
Jul 28 at 15:20
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
$$y = ln (|ln x|)$$
Attempt at solution:
For the function to be defined, $|ln x|> 0$
$implies ln x > 0 $
$implies x > 1$ [$because$ anti-logging both sides.]
But clearly, the function is valid if $xin(0,1)$ as well.
I am not sure what I am doing wrong. I might be doing something completely illegal here. So sorry about that!
I am looking for a mechanical way of solving these types of problems. So solutions which don't involve graphs (if possible) are requested.
Thanks in advance!
Ps. I have no clue on how to begin calculating for the range.
logarithms graphing-functions absolute-value
asked Jul 28 at 11:11
Amaimon43
11
$$y = ln (|ln x|)$$
Attempt at solution:
For the function to be defined, $|ln x|> 0$
$implies ln x > 0 $
$implies x > 1$ [$because$ anti-logging both sides.]
But clearly, the function is valid if $xin(0,1)$ as well.
I am not sure what I am doing wrong. I might be doing something completely illegal here. So sorry about that!
I am looking for a mechanical way of solving these types of problems. So solutions which don't involve graphs (if possible) are requested.
Thanks in advance!
Ps. I have no clue on how to begin calculating for the range.
logarithms graphing-functions absolute-value
asked Jul 28 at 11:11
Amaimon43
11
edited Jul 28 at 15:20
asked Jul 28 at 11:11
Amaimon43
11
asked Jul 28 at 11:11
Amaimon43
11
asked Jul 28 at 11:11
Amaimon43
11
11
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 28 at 11:25
With $mod$ do you mean the absolute value?
– Davide Morgante
Jul 28 at 11:39
Ya that's what I meant.
– Amaimon43
Jul 28 at 15:20
add a comment |Â
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 28 at 11:25
With $mod$ do you mean the absolute value?
– Davide Morgante
Jul 28 at 11:39
Ya that's what I meant.
– Amaimon43
Jul 28 at 15:20
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 28 at 11:25
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 28 at 11:25
With $mod$ do you mean the absolute value?
– Davide Morgante
Jul 28 at 11:39
With $mod$ do you mean the absolute value?
– Davide Morgante
Jul 28 at 11:39
Ya that's what I meant.
– Amaimon43
Jul 28 at 15:20
Ya that's what I meant.
– Amaimon43
Jul 28 at 15:20
add a comment |Â
1 Answer
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As you observed, $y=ln(x)$ is defined for $x>0$. Notice that $|ln(x)|$ is always greater than $0$, except at $x=1$, where $|ln(x)|=0$. Therefore, we can take all $x>0$ where $xneq 1$, and there will be a vertical asymptote at $x=1$ which "splits" the function into two pieces. Together, their domain is $D=lbrace x>0 |xneq 1rbrace$ and the range is $R=mathbbR$. Just looking at the first "piece" of the function with domain $D_1=lbrace xinmathbbR |0<x<1rbrace$, we can clearly see that the range is all reals. This is because the function $ln(x)$ restricted to the domain $D_1$ has range $R_1=lbrace xinmathbbR|-infty<x<0rbrace$, so $|ln(x)|$ has range $R_2=lbrace xinmathbbR|0<x<inftyrbrace$ which is the original domain of the natural log function!
answered Jul 28 at 15:48
高田航
1,116318
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
As you observed, $y=ln(x)$ is defined for $x>0$. Notice that $|ln(x)|$ is always greater than $0$, except at $x=1$, where $|ln(x)|=0$. Therefore, we can take all $x>0$ where $xneq 1$, and there will be a vertical asymptote at $x=1$ which "splits" the function into two pieces. Together, their domain is $D=lbrace x>0 |xneq 1rbrace$ and the range is $R=mathbbR$. Just looking at the first "piece" of the function with domain $D_1=lbrace xinmathbbR |0<x<1rbrace$, we can clearly see that the range is all reals. This is because the function $ln(x)$ restricted to the domain $D_1$ has range $R_1=lbrace xinmathbbR|-infty<x<0rbrace$, so $|ln(x)|$ has range $R_2=lbrace xinmathbbR|0<x<inftyrbrace$ which is the original domain of the natural log function!
answered Jul 28 at 15:48
高田航
1,116318
add a comment |Â
up vote
0
down vote
As you observed, $y=ln(x)$ is defined for $x>0$. Notice that $|ln(x)|$ is always greater than $0$, except at $x=1$, where $|ln(x)|=0$. Therefore, we can take all $x>0$ where $xneq 1$, and there will be a vertical asymptote at $x=1$ which "splits" the function into two pieces. Together, their domain is $D=lbrace x>0 |xneq 1rbrace$ and the range is $R=mathbbR$. Just looking at the first "piece" of the function with domain $D_1=lbrace xinmathbbR |0<x<1rbrace$, we can clearly see that the range is all reals. This is because the function $ln(x)$ restricted to the domain $D_1$ has range $R_1=lbrace xinmathbbR|-infty<x<0rbrace$, so $|ln(x)|$ has range $R_2=lbrace xinmathbbR|0<x<inftyrbrace$ which is the original domain of the natural log function!
answered Jul 28 at 15:48
高田航
1,116318
add a comment |Â
up vote
0
down vote
up vote
0
down vote
As you observed, $y=ln(x)$ is defined for $x>0$. Notice that $|ln(x)|$ is always greater than $0$, except at $x=1$, where $|ln(x)|=0$. Therefore, we can take all $x>0$ where $xneq 1$, and there will be a vertical asymptote at $x=1$ which "splits" the function into two pieces. Together, their domain is $D=lbrace x>0 |xneq 1rbrace$ and the range is $R=mathbbR$. Just looking at the first "piece" of the function with domain $D_1=lbrace xinmathbbR |0<x<1rbrace$, we can clearly see that the range is all reals. This is because the function $ln(x)$ restricted to the domain $D_1$ has range $R_1=lbrace xinmathbbR|-infty<x<0rbrace$, so $|ln(x)|$ has range $R_2=lbrace xinmathbbR|0<x<inftyrbrace$ which is the original domain of the natural log function!
answered Jul 28 at 15:48
高田航
1,116318
As you observed, $y=ln(x)$ is defined for $x>0$. Notice that $|ln(x)|$ is always greater than $0$, except at $x=1$, where $|ln(x)|=0$. Therefore, we can take all $x>0$ where $xneq 1$, and there will be a vertical asymptote at $x=1$ which "splits" the function into two pieces. Together, their domain is $D=lbrace x>0 |xneq 1rbrace$ and the range is $R=mathbbR$. Just looking at the first "piece" of the function with domain $D_1=lbrace xinmathbbR |0<x<1rbrace$, we can clearly see that the range is all reals. This is because the function $ln(x)$ restricted to the domain $D_1$ has range $R_1=lbrace xinmathbbR|-infty<x<0rbrace$, so $|ln(x)|$ has range $R_2=lbrace xinmathbbR|0<x<inftyrbrace$ which is the original domain of the natural log function!
answered Jul 28 at 15:48
高田航
1,116318
answered Jul 28 at 15:48
高田航
1,116318
answered Jul 28 at 15:48
高田航
1,116318
answered Jul 28 at 15:48
高田航
1,116318
1,116318
add a comment |Â
add a comment |Â
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Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 28 at 11:25
With $mod$ do you mean the absolute value?
– Davide Morgante
Jul 28 at 11:39
Ya that's what I meant.
– Amaimon43
Jul 28 at 15:20