Why does this ln function inverse?
Multi tool use
Clash Royale CLAN TAG#URR8PPP
up vote
5
down vote
favorite
I'm going through applications of separable equations and came across an example of half-lives:
$$M(t)=fracM2=Me^-kt$$
Factoring out $M$, $frac12=e^-kt$.
To solve for $t$, $lnleft(frac12right)=-kt$.
And then the answer is $t=frac1kln 2$.
Why isn't the value for $t=-frac1kln frac12$?
Thanks:)
calculus logarithms
edited Aug 31 at 11:18
Carsten S
6,72411334
asked Aug 31 at 8:37
Hews
515
add a comment |Â
up vote
5
down vote
favorite
I'm going through applications of separable equations and came across an example of half-lives:
$$M(t)=fracM2=Me^-kt$$
Factoring out $M$, $frac12=e^-kt$.
To solve for $t$, $lnleft(frac12right)=-kt$.
And then the answer is $t=frac1kln 2$.
Why isn't the value for $t=-frac1kln frac12$?
Thanks:)
calculus logarithms
edited Aug 31 at 11:18
Carsten S
6,72411334
asked Aug 31 at 8:37
Hews
515
12
It is the same !
– Claude Leibovici
Aug 31 at 8:42
1
Recall the log sum rule: $log(x) + log(y) =log(xy)$. Apply this with $y=1/x$ to derive $log(x) = -log(1/x)$.
– Winther
Aug 31 at 8:44
1
Also, to complement @Winther's comment, note that $log x^n = n log x$ and $x^-1=frac1x$.
– User123456789
Aug 31 at 11:30
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
I'm going through applications of separable equations and came across an example of half-lives:
$$M(t)=fracM2=Me^-kt$$
Factoring out $M$, $frac12=e^-kt$.
To solve for $t$, $lnleft(frac12right)=-kt$.
And then the answer is $t=frac1kln 2$.
Why isn't the value for $t=-frac1kln frac12$?
Thanks:)
calculus logarithms
edited Aug 31 at 11:18
Carsten S
6,72411334
asked Aug 31 at 8:37
Hews
515
I'm going through applications of separable equations and came across an example of half-lives:
$$M(t)=fracM2=Me^-kt$$
Factoring out $M$, $frac12=e^-kt$.
To solve for $t$, $lnleft(frac12right)=-kt$.
And then the answer is $t=frac1kln 2$.
Why isn't the value for $t=-frac1kln frac12$?
Thanks:)
calculus logarithms
calculus logarithms
edited Aug 31 at 11:18
Carsten S
6,72411334
asked Aug 31 at 8:37
Hews
515
edited Aug 31 at 11:18
Carsten S
6,72411334
asked Aug 31 at 8:37
Hews
515
edited Aug 31 at 11:18
Carsten S
6,72411334
edited Aug 31 at 11:18
Carsten S
6,72411334
edited Aug 31 at 11:18
Carsten S
6,72411334
6,72411334
asked Aug 31 at 8:37
Hews
515
asked Aug 31 at 8:37
Hews
515
asked Aug 31 at 8:37
Hews
515
515
12
It is the same !
– Claude Leibovici
Aug 31 at 8:42
1
Recall the log sum rule: $log(x) + log(y) =log(xy)$. Apply this with $y=1/x$ to derive $log(x) = -log(1/x)$.
– Winther
Aug 31 at 8:44
1
Also, to complement @Winther's comment, note that $log x^n = n log x$ and $x^-1=frac1x$.
– User123456789
Aug 31 at 11:30
add a comment |Â
12
It is the same !
– Claude Leibovici
Aug 31 at 8:42
1
Recall the log sum rule: $log(x) + log(y) =log(xy)$. Apply this with $y=1/x$ to derive $log(x) = -log(1/x)$.
– Winther
Aug 31 at 8:44
1
Also, to complement @Winther's comment, note that $log x^n = n log x$ and $x^-1=frac1x$.
– User123456789
Aug 31 at 11:30
12
12
It is the same !
– Claude Leibovici
Aug 31 at 8:42
It is the same !
– Claude Leibovici
Aug 31 at 8:42
1
1
Recall the log sum rule: $log(x) + log(y) =log(xy)$. Apply this with $y=1/x$ to derive $log(x) = -log(1/x)$.
– Winther
Aug 31 at 8:44
Recall the log sum rule: $log(x) + log(y) =log(xy)$. Apply this with $y=1/x$ to derive $log(x) = -log(1/x)$.
– Winther
Aug 31 at 8:44
1
1
Also, to complement @Winther's comment, note that $log x^n = n log x$ and $x^-1=frac1x$.
– User123456789
Aug 31 at 11:30
Also, to complement @Winther's comment, note that $log x^n = n log x$ and $x^-1=frac1x$.
– User123456789
Aug 31 at 11:30
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
12
down vote
accepted
It is the same since $$ln(2)=-ln(1/2).$$
answered Aug 31 at 8:42
Hello_World
3,40021429
add a comment |Â
up vote
4
down vote
$ln(1/2) = -kt$ implies $-ln(2) = -kt$ implies $dfracln(2)k=t$ which is also $-dfracln(1/2)k=t$.
Reminder that $ln(1/x) = -ln(x)$.
edited Aug 31 at 12:56
BPP
1,999827
answered Aug 31 at 8:43
PackSciences
41616
add a comment |Â
up vote
3
down vote
$lnfrac12 = ln1 - ln2 = 0 - ln2 = -ln2$
answered Aug 31 at 15:10
Florian
312
@Leucippus I disagree. This is a legitimate answer to the question.
– Brahadeesh
Aug 31 at 15:53
add a comment |Â
up vote
0
down vote
Given the exponent rule for logarithms $log(a^b) = blog(a)$ with $a = 1/2$ and $b = -1$ we have
$$
log(2) = log((1/2)^-1) = -log(1/2)
$$
answered Aug 31 at 13:44
Frank Vel
2,32942242
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
12
down vote
accepted
It is the same since $$ln(2)=-ln(1/2).$$
answered Aug 31 at 8:42
Hello_World
3,40021429
add a comment |Â
up vote
12
down vote
accepted
It is the same since $$ln(2)=-ln(1/2).$$
answered Aug 31 at 8:42
Hello_World
3,40021429
add a comment |Â
up vote
12
down vote
accepted
up vote
12
down vote
accepted
It is the same since $$ln(2)=-ln(1/2).$$
answered Aug 31 at 8:42
Hello_World
3,40021429
It is the same since $$ln(2)=-ln(1/2).$$
answered Aug 31 at 8:42
Hello_World
3,40021429
answered Aug 31 at 8:42
Hello_World
3,40021429
answered Aug 31 at 8:42
Hello_World
3,40021429
answered Aug 31 at 8:42
Hello_World
3,40021429
3,40021429
add a comment |Â
add a comment |Â
up vote
4
down vote
$ln(1/2) = -kt$ implies $-ln(2) = -kt$ implies $dfracln(2)k=t$ which is also $-dfracln(1/2)k=t$.
Reminder that $ln(1/x) = -ln(x)$.
edited Aug 31 at 12:56
BPP
1,999827
answered Aug 31 at 8:43
PackSciences
41616
add a comment |Â
up vote
4
down vote
$ln(1/2) = -kt$ implies $-ln(2) = -kt$ implies $dfracln(2)k=t$ which is also $-dfracln(1/2)k=t$.
Reminder that $ln(1/x) = -ln(x)$.
edited Aug 31 at 12:56
BPP
1,999827
answered Aug 31 at 8:43
PackSciences
41616
add a comment |Â
up vote
4
down vote
up vote
4
down vote
$ln(1/2) = -kt$ implies $-ln(2) = -kt$ implies $dfracln(2)k=t$ which is also $-dfracln(1/2)k=t$.
Reminder that $ln(1/x) = -ln(x)$.
edited Aug 31 at 12:56
BPP
1,999827
answered Aug 31 at 8:43
PackSciences
41616
$ln(1/2) = -kt$ implies $-ln(2) = -kt$ implies $dfracln(2)k=t$ which is also $-dfracln(1/2)k=t$.
Reminder that $ln(1/x) = -ln(x)$.
edited Aug 31 at 12:56
BPP
1,999827
answered Aug 31 at 8:43
PackSciences
41616
edited Aug 31 at 12:56
BPP
1,999827
edited Aug 31 at 12:56
BPP
1,999827
edited Aug 31 at 12:56
BPP
1,999827
1,999827
answered Aug 31 at 8:43
PackSciences
41616
answered Aug 31 at 8:43
PackSciences
41616
answered Aug 31 at 8:43
PackSciences
41616
41616
add a comment |Â
add a comment |Â
up vote
3
down vote
$lnfrac12 = ln1 - ln2 = 0 - ln2 = -ln2$
answered Aug 31 at 15:10
Florian
312
@Leucippus I disagree. This is a legitimate answer to the question.
– Brahadeesh
Aug 31 at 15:53
add a comment |Â
up vote
3
down vote
$lnfrac12 = ln1 - ln2 = 0 - ln2 = -ln2$
answered Aug 31 at 15:10
Florian
312
@Leucippus I disagree. This is a legitimate answer to the question.
– Brahadeesh
Aug 31 at 15:53
add a comment |Â
up vote
3
down vote
up vote
3
down vote
$lnfrac12 = ln1 - ln2 = 0 - ln2 = -ln2$
answered Aug 31 at 15:10
Florian
312
$lnfrac12 = ln1 - ln2 = 0 - ln2 = -ln2$
answered Aug 31 at 15:10
Florian
312
answered Aug 31 at 15:10
Florian
312
answered Aug 31 at 15:10
Florian
312
answered Aug 31 at 15:10
Florian
312
312
@Leucippus I disagree. This is a legitimate answer to the question.
– Brahadeesh
Aug 31 at 15:53
add a comment |Â
@Leucippus I disagree. This is a legitimate answer to the question.
– Brahadeesh
Aug 31 at 15:53
@Leucippus I disagree. This is a legitimate answer to the question.
– Brahadeesh
Aug 31 at 15:53
@Leucippus I disagree. This is a legitimate answer to the question.
– Brahadeesh
Aug 31 at 15:53
add a comment |Â
up vote
0
down vote
Given the exponent rule for logarithms $log(a^b) = blog(a)$ with $a = 1/2$ and $b = -1$ we have
$$
log(2) = log((1/2)^-1) = -log(1/2)
$$
answered Aug 31 at 13:44
Frank Vel
2,32942242
add a comment |Â
up vote
0
down vote
Given the exponent rule for logarithms $log(a^b) = blog(a)$ with $a = 1/2$ and $b = -1$ we have
$$
log(2) = log((1/2)^-1) = -log(1/2)
$$
answered Aug 31 at 13:44
Frank Vel
2,32942242
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Given the exponent rule for logarithms $log(a^b) = blog(a)$ with $a = 1/2$ and $b = -1$ we have
$$
log(2) = log((1/2)^-1) = -log(1/2)
$$
answered Aug 31 at 13:44
Frank Vel
2,32942242
Given the exponent rule for logarithms $log(a^b) = blog(a)$ with $a = 1/2$ and $b = -1$ we have
$$
log(2) = log((1/2)^-1) = -log(1/2)
$$
answered Aug 31 at 13:44
Frank Vel
2,32942242
answered Aug 31 at 13:44
Frank Vel
2,32942242
answered Aug 31 at 13:44
Frank Vel
2,32942242
answered Aug 31 at 13:44
Frank Vel
2,32942242
2,32942242
add a comment |Â
add a comment |Â
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2900468%2fwhy-does-this-ln-function-inverse%23new-answer', 'question_page');
);
Post as a guest
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
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
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
12
It is the same !
– Claude Leibovici
Aug 31 at 8:42
1
Recall the log sum rule: $log(x) + log(y) =log(xy)$. Apply this with $y=1/x$ to derive $log(x) = -log(1/x)$.
– Winther
Aug 31 at 8:44
1
Also, to complement @Winther's comment, note that $log x^n = n log x$ and $x^-1=frac1x$.
– User123456789
Aug 31 at 11:30