Mixing
$(frac{3}{2})^h = log_{3/2} h = log_{2} h / log_{2} 3/2$
$begingroup$
$(frac{3}{2})^h$, then take the log of it: $log_{3/2} h = log_{2} h / log_{2} 3/2$
could someone explain steps please.
I thought $(3/2)^h$ would be $hlog (3/2)$
nor do I get how they got to this $log_{3/2} h = log_{2} h / log_{2} 3/2$
Which property are they optimizing?
algebra-precalculus
asked Feb 2 at 2:03
Bas basBas bas
49512
$endgroup$
|
show 2 more comments
$begingroup$
$(frac{3}{2})^h$, then take the log of it: $log_{3/2} h = log_{2} h / log_{2} 3/2$
could someone explain steps please.
I thought $(3/2)^h$ would be $hlog (3/2)$
nor do I get how they got to this $log_{3/2} h = log_{2} h / log_{2} 3/2$
Which property are they optimizing?
algebra-precalculus
asked Feb 2 at 2:03
Bas basBas bas
49512
$endgroup$
$begingroup$
For me also, the first line doesn't seem to make sense. Also, what is $n$? Where does this come from?
$endgroup$
– John Omielan
Feb 2 at 2:10
$begingroup$
n should be h - whoops
$endgroup$
– Bas bas
Feb 2 at 2:12
$begingroup$
Thanks for the feedback & update of the question text. You should also update the title, including removing the first "=" as that is not true. In addition, please let us know where the first line comes from. Thanks.
$endgroup$
– John Omielan
Feb 2 at 2:13
$begingroup$
This question makes no sense. What exactly are you asking? What is the context?
$endgroup$
– Zubin Mukerjee
Feb 2 at 2:22
$begingroup$
How does $log_{3/2} h = log_{2} h / log_{2} 3/2$. What properties are used?
$endgroup$
– Bas bas
Feb 2 at 2:24
|
show 2 more comments
$begingroup$
$(frac{3}{2})^h$, then take the log of it: $log_{3/2} h = log_{2} h / log_{2} 3/2$
could someone explain steps please.
I thought $(3/2)^h$ would be $hlog (3/2)$
nor do I get how they got to this $log_{3/2} h = log_{2} h / log_{2} 3/2$
Which property are they optimizing?
algebra-precalculus
asked Feb 2 at 2:03
Bas basBas bas
49512
$endgroup$
$(frac{3}{2})^h$, then take the log of it: $log_{3/2} h = log_{2} h / log_{2} 3/2$
could someone explain steps please.
I thought $(3/2)^h$ would be $hlog (3/2)$
nor do I get how they got to this $log_{3/2} h = log_{2} h / log_{2} 3/2$
Which property are they optimizing?
algebra-precalculus
algebra-precalculus
asked Feb 2 at 2:03
Bas basBas bas
49512
asked Feb 2 at 2:03
Bas basBas bas
49512
edited Feb 2 at 2:25
Bas bas
asked Feb 2 at 2:03
Bas basBas bas
49512
asked Feb 2 at 2:03
Bas basBas bas
49512
asked Feb 2 at 2:03
Bas basBas bas
49512
49512
$begingroup$
For me also, the first line doesn't seem to make sense. Also, what is $n$? Where does this come from?
$endgroup$
– John Omielan
Feb 2 at 2:10
$begingroup$
n should be h - whoops
$endgroup$
– Bas bas
Feb 2 at 2:12
$begingroup$
Thanks for the feedback & update of the question text. You should also update the title, including removing the first "=" as that is not true. In addition, please let us know where the first line comes from. Thanks.
$endgroup$
– John Omielan
Feb 2 at 2:13
$begingroup$
This question makes no sense. What exactly are you asking? What is the context?
$endgroup$
– Zubin Mukerjee
Feb 2 at 2:22
$begingroup$
How does $log_{3/2} h = log_{2} h / log_{2} 3/2$. What properties are used?
$endgroup$
– Bas bas
Feb 2 at 2:24
|
show 2 more comments
$begingroup$
For me also, the first line doesn't seem to make sense. Also, what is $n$? Where does this come from?
$endgroup$
– John Omielan
Feb 2 at 2:10
$begingroup$
n should be h - whoops
$endgroup$
– Bas bas
Feb 2 at 2:12
$begingroup$
Thanks for the feedback & update of the question text. You should also update the title, including removing the first "=" as that is not true. In addition, please let us know where the first line comes from. Thanks.
$endgroup$
– John Omielan
Feb 2 at 2:13
$begingroup$
This question makes no sense. What exactly are you asking? What is the context?
$endgroup$
– Zubin Mukerjee
Feb 2 at 2:22
$begingroup$
How does $log_{3/2} h = log_{2} h / log_{2} 3/2$. What properties are used?
$endgroup$
– Bas bas
Feb 2 at 2:24
$begingroup$
For me also, the first line doesn't seem to make sense. Also, what is $n$? Where does this come from?
$endgroup$
– John Omielan
Feb 2 at 2:10
$begingroup$
For me also, the first line doesn't seem to make sense. Also, what is $n$? Where does this come from?
$endgroup$
– John Omielan
Feb 2 at 2:10
$begingroup$
n should be h - whoops
$endgroup$
– Bas bas
Feb 2 at 2:12
$begingroup$
n should be h - whoops
$endgroup$
– Bas bas
Feb 2 at 2:12
$begingroup$
Thanks for the feedback & update of the question text. You should also update the title, including removing the first "=" as that is not true. In addition, please let us know where the first line comes from. Thanks.
$endgroup$
– John Omielan
Feb 2 at 2:13
$begingroup$
Thanks for the feedback & update of the question text. You should also update the title, including removing the first "=" as that is not true. In addition, please let us know where the first line comes from. Thanks.
$endgroup$
– John Omielan
Feb 2 at 2:13
$begingroup$
This question makes no sense. What exactly are you asking? What is the context?
$endgroup$
– Zubin Mukerjee
Feb 2 at 2:22
$begingroup$
This question makes no sense. What exactly are you asking? What is the context?
$endgroup$
– Zubin Mukerjee
Feb 2 at 2:22
$begingroup$
How does $log_{3/2} h = log_{2} h / log_{2} 3/2$. What properties are used?
$endgroup$
– Bas bas
Feb 2 at 2:24
$begingroup$
How does $log_{3/2} h = log_{2} h / log_{2} 3/2$. What properties are used?
$endgroup$
– Bas bas
Feb 2 at 2:24
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
If you take the (natural) log of $left(frac{3}{2}right)^h$, then it is indeed $hlogleft(frac{3}{2}right)$.
It appears that whatever source you got this from attempted to take the log base $frac{3}{2}$ of $left(frac{3}{2}right)^h$. But this would be $log_{3/2}left(left(frac{3}{2}right)^hright)=hlog_{3/2}left(frac{3}{2}right)=h$, not $log_{3/2}(h)$.
One rule of logarithms is that $log_a(b) = frac{log_c(b)}{log_c(a)}$ (this is the change of base formula). Applying this with $a=frac{3}{2}$, $b=h$, and $c=2$, we do indeed have $log_{3/2}(h) = log_2(h)/log_2(3/2)$.
In conclusion,
$logleft(left(frac{3}{2}right)^hright)neq log_{3/2}(h)$ (in any base)
$log_{3/2}(h) = log_2(h)/log_2(3/2)$ by the change of base formula.
answered Feb 2 at 2:25
kccukccu
11.2k11231
$endgroup$
$begingroup$
As such, it seems the $log_{3/2}{left(hright)} = log_{2}{left(hright)} / log_{2}{left(3/2right)}$ equation is true for any positive $h$. Thus, it doesn't really have anything directly to do with the original value of $left(frac{3}{2}right)^h$.
$endgroup$
– John Omielan
Feb 2 at 2:31
$begingroup$
That is correct, the value of $h$ is irrelevant (well, it has to be positive to take the log of it, but that's it).
$endgroup$
– kccu
Feb 2 at 2:32
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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active
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active
oldest
votes
$begingroup$
If you take the (natural) log of $left(frac{3}{2}right)^h$, then it is indeed $hlogleft(frac{3}{2}right)$.
It appears that whatever source you got this from attempted to take the log base $frac{3}{2}$ of $left(frac{3}{2}right)^h$. But this would be $log_{3/2}left(left(frac{3}{2}right)^hright)=hlog_{3/2}left(frac{3}{2}right)=h$, not $log_{3/2}(h)$.
One rule of logarithms is that $log_a(b) = frac{log_c(b)}{log_c(a)}$ (this is the change of base formula). Applying this with $a=frac{3}{2}$, $b=h$, and $c=2$, we do indeed have $log_{3/2}(h) = log_2(h)/log_2(3/2)$.
In conclusion,
$logleft(left(frac{3}{2}right)^hright)neq log_{3/2}(h)$ (in any base)
$log_{3/2}(h) = log_2(h)/log_2(3/2)$ by the change of base formula.
answered Feb 2 at 2:25
kccukccu
11.2k11231
$endgroup$
$begingroup$
As such, it seems the $log_{3/2}{left(hright)} = log_{2}{left(hright)} / log_{2}{left(3/2right)}$ equation is true for any positive $h$. Thus, it doesn't really have anything directly to do with the original value of $left(frac{3}{2}right)^h$.
$endgroup$
– John Omielan
Feb 2 at 2:31
$begingroup$
That is correct, the value of $h$ is irrelevant (well, it has to be positive to take the log of it, but that's it).
$endgroup$
– kccu
Feb 2 at 2:32
add a comment |
$begingroup$
If you take the (natural) log of $left(frac{3}{2}right)^h$, then it is indeed $hlogleft(frac{3}{2}right)$.
It appears that whatever source you got this from attempted to take the log base $frac{3}{2}$ of $left(frac{3}{2}right)^h$. But this would be $log_{3/2}left(left(frac{3}{2}right)^hright)=hlog_{3/2}left(frac{3}{2}right)=h$, not $log_{3/2}(h)$.
One rule of logarithms is that $log_a(b) = frac{log_c(b)}{log_c(a)}$ (this is the change of base formula). Applying this with $a=frac{3}{2}$, $b=h$, and $c=2$, we do indeed have $log_{3/2}(h) = log_2(h)/log_2(3/2)$.
In conclusion,
$logleft(left(frac{3}{2}right)^hright)neq log_{3/2}(h)$ (in any base)
$log_{3/2}(h) = log_2(h)/log_2(3/2)$ by the change of base formula.
answered Feb 2 at 2:25
kccukccu
11.2k11231
$endgroup$
$begingroup$
As such, it seems the $log_{3/2}{left(hright)} = log_{2}{left(hright)} / log_{2}{left(3/2right)}$ equation is true for any positive $h$. Thus, it doesn't really have anything directly to do with the original value of $left(frac{3}{2}right)^h$.
$endgroup$
– John Omielan
Feb 2 at 2:31
$begingroup$
That is correct, the value of $h$ is irrelevant (well, it has to be positive to take the log of it, but that's it).
$endgroup$
– kccu
Feb 2 at 2:32
add a comment |
$begingroup$
If you take the (natural) log of $left(frac{3}{2}right)^h$, then it is indeed $hlogleft(frac{3}{2}right)$.
It appears that whatever source you got this from attempted to take the log base $frac{3}{2}$ of $left(frac{3}{2}right)^h$. But this would be $log_{3/2}left(left(frac{3}{2}right)^hright)=hlog_{3/2}left(frac{3}{2}right)=h$, not $log_{3/2}(h)$.
One rule of logarithms is that $log_a(b) = frac{log_c(b)}{log_c(a)}$ (this is the change of base formula). Applying this with $a=frac{3}{2}$, $b=h$, and $c=2$, we do indeed have $log_{3/2}(h) = log_2(h)/log_2(3/2)$.
In conclusion,
$logleft(left(frac{3}{2}right)^hright)neq log_{3/2}(h)$ (in any base)
$log_{3/2}(h) = log_2(h)/log_2(3/2)$ by the change of base formula.
answered Feb 2 at 2:25
kccukccu
11.2k11231
$endgroup$
If you take the (natural) log of $left(frac{3}{2}right)^h$, then it is indeed $hlogleft(frac{3}{2}right)$.
It appears that whatever source you got this from attempted to take the log base $frac{3}{2}$ of $left(frac{3}{2}right)^h$. But this would be $log_{3/2}left(left(frac{3}{2}right)^hright)=hlog_{3/2}left(frac{3}{2}right)=h$, not $log_{3/2}(h)$.
One rule of logarithms is that $log_a(b) = frac{log_c(b)}{log_c(a)}$ (this is the change of base formula). Applying this with $a=frac{3}{2}$, $b=h$, and $c=2$, we do indeed have $log_{3/2}(h) = log_2(h)/log_2(3/2)$.
In conclusion,
$logleft(left(frac{3}{2}right)^hright)neq log_{3/2}(h)$ (in any base)
$log_{3/2}(h) = log_2(h)/log_2(3/2)$ by the change of base formula.
answered Feb 2 at 2:25
kccukccu
11.2k11231
answered Feb 2 at 2:25
kccukccu
11.2k11231
answered Feb 2 at 2:25
kccukccu
11.2k11231
answered Feb 2 at 2:25
kccukccu
11.2k11231
11.2k11231
$begingroup$
As such, it seems the $log_{3/2}{left(hright)} = log_{2}{left(hright)} / log_{2}{left(3/2right)}$ equation is true for any positive $h$. Thus, it doesn't really have anything directly to do with the original value of $left(frac{3}{2}right)^h$.
$endgroup$
– John Omielan
Feb 2 at 2:31
$begingroup$
That is correct, the value of $h$ is irrelevant (well, it has to be positive to take the log of it, but that's it).
$endgroup$
– kccu
Feb 2 at 2:32
add a comment |
$begingroup$
As such, it seems the $log_{3/2}{left(hright)} = log_{2}{left(hright)} / log_{2}{left(3/2right)}$ equation is true for any positive $h$. Thus, it doesn't really have anything directly to do with the original value of $left(frac{3}{2}right)^h$.
$endgroup$
– John Omielan
Feb 2 at 2:31
$begingroup$
That is correct, the value of $h$ is irrelevant (well, it has to be positive to take the log of it, but that's it).
$endgroup$
– kccu
Feb 2 at 2:32
$begingroup$
As such, it seems the $log_{3/2}{left(hright)} = log_{2}{left(hright)} / log_{2}{left(3/2right)}$ equation is true for any positive $h$. Thus, it doesn't really have anything directly to do with the original value of $left(frac{3}{2}right)^h$.
$endgroup$
– John Omielan
Feb 2 at 2:31
$begingroup$
As such, it seems the $log_{3/2}{left(hright)} = log_{2}{left(hright)} / log_{2}{left(3/2right)}$ equation is true for any positive $h$. Thus, it doesn't really have anything directly to do with the original value of $left(frac{3}{2}right)^h$.
$endgroup$
– John Omielan
Feb 2 at 2:31
$begingroup$
That is correct, the value of $h$ is irrelevant (well, it has to be positive to take the log of it, but that's it).
$endgroup$
– kccu
Feb 2 at 2:32
$begingroup$
That is correct, the value of $h$ is irrelevant (well, it has to be positive to take the log of it, but that's it).
$endgroup$
– kccu
Feb 2 at 2:32
add a comment |
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$begingroup$
For me also, the first line doesn't seem to make sense. Also, what is $n$? Where does this come from?
$endgroup$
– John Omielan
Feb 2 at 2:10
$begingroup$
n should be h - whoops
$endgroup$
– Bas bas
Feb 2 at 2:12
$begingroup$
Thanks for the feedback & update of the question text. You should also update the title, including removing the first "=" as that is not true. In addition, please let us know where the first line comes from. Thanks.
$endgroup$
– John Omielan
Feb 2 at 2:13
$begingroup$
This question makes no sense. What exactly are you asking? What is the context?
$endgroup$
– Zubin Mukerjee
Feb 2 at 2:22
$begingroup$
How does $log_{3/2} h = log_{2} h / log_{2} 3/2$. What properties are used?
$endgroup$
– Bas bas
Feb 2 at 2:24