Why does this ln function inverse?
Clash Royale CLAN TAG#URR8PPP
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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
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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
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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