Mixing
Exponents Linear algebra
$begingroup$
$x^y=z $
Proof that
$x^n/z=y$
I was calculating the cube root of $2$ by hand and when checking it out, I noticed its square is close to the value of logarithm of $3$ in base $2$. A little tweaking and I got the exact value for $n$ when $z$ is set at $3$. I wonder if there is detailed proof of such operations. Thanks.
logarithms
edited Jan 29 at 21:29
abc...
3,237739
asked Jan 29 at 21:23
Adam BoitAdam Boit
164
$endgroup$
add a comment |
$begingroup$
$x^y=z $
Proof that
$x^n/z=y$
I was calculating the cube root of $2$ by hand and when checking it out, I noticed its square is close to the value of logarithm of $3$ in base $2$. A little tweaking and I got the exact value for $n$ when $z$ is set at $3$. I wonder if there is detailed proof of such operations. Thanks.
logarithms
edited Jan 29 at 21:29
abc...
3,237739
asked Jan 29 at 21:23
Adam BoitAdam Boit
164
$endgroup$
1
$begingroup$
What is n in this case?
$endgroup$
– Gnumbertester
Jan 29 at 21:30
$begingroup$
n/z = 1.99334613/3
$endgroup$
– Adam Boit
Jan 29 at 22:39
$begingroup$
What do $n$ and $z$ actually represent though?
$endgroup$
– Gnumbertester
Jan 29 at 22:46
$begingroup$
n and z is fractional exponent where the index of the root just happens to equal the number we get when we make y the new exponent.
$endgroup$
– Adam Boit
Jan 30 at 5:12
add a comment |
$begingroup$
$x^y=z $
Proof that
$x^n/z=y$
I was calculating the cube root of $2$ by hand and when checking it out, I noticed its square is close to the value of logarithm of $3$ in base $2$. A little tweaking and I got the exact value for $n$ when $z$ is set at $3$. I wonder if there is detailed proof of such operations. Thanks.
logarithms
edited Jan 29 at 21:29
abc...
3,237739
asked Jan 29 at 21:23
Adam BoitAdam Boit
164
$endgroup$
$x^y=z $
Proof that
$x^n/z=y$
I was calculating the cube root of $2$ by hand and when checking it out, I noticed its square is close to the value of logarithm of $3$ in base $2$. A little tweaking and I got the exact value for $n$ when $z$ is set at $3$. I wonder if there is detailed proof of such operations. Thanks.
logarithms
logarithms
edited Jan 29 at 21:29
abc...
3,237739
asked Jan 29 at 21:23
Adam BoitAdam Boit
164
edited Jan 29 at 21:29
abc...
3,237739
asked Jan 29 at 21:23
Adam BoitAdam Boit
164
edited Jan 29 at 21:29
abc...
3,237739
edited Jan 29 at 21:29
abc...
3,237739
edited Jan 29 at 21:29
abc...
3,237739
3,237739
asked Jan 29 at 21:23
Adam BoitAdam Boit
164
asked Jan 29 at 21:23
Adam BoitAdam Boit
164
asked Jan 29 at 21:23
Adam BoitAdam Boit
164
164
1
$begingroup$
What is n in this case?
$endgroup$
– Gnumbertester
Jan 29 at 21:30
$begingroup$
n/z = 1.99334613/3
$endgroup$
– Adam Boit
Jan 29 at 22:39
$begingroup$
What do $n$ and $z$ actually represent though?
$endgroup$
– Gnumbertester
Jan 29 at 22:46
$begingroup$
n and z is fractional exponent where the index of the root just happens to equal the number we get when we make y the new exponent.
$endgroup$
– Adam Boit
Jan 30 at 5:12
add a comment |
1
$begingroup$
What is n in this case?
$endgroup$
– Gnumbertester
Jan 29 at 21:30
$begingroup$
n/z = 1.99334613/3
$endgroup$
– Adam Boit
Jan 29 at 22:39
$begingroup$
What do $n$ and $z$ actually represent though?
$endgroup$
– Gnumbertester
Jan 29 at 22:46
$begingroup$
n and z is fractional exponent where the index of the root just happens to equal the number we get when we make y the new exponent.
$endgroup$
– Adam Boit
Jan 30 at 5:12
1
1
$begingroup$
What is n in this case?
$endgroup$
– Gnumbertester
Jan 29 at 21:30
$begingroup$
What is n in this case?
$endgroup$
– Gnumbertester
Jan 29 at 21:30
$begingroup$
n/z = 1.99334613/3
$endgroup$
– Adam Boit
Jan 29 at 22:39
$begingroup$
n/z = 1.99334613/3
$endgroup$
– Adam Boit
Jan 29 at 22:39
$begingroup$
What do $n$ and $z$ actually represent though?
$endgroup$
– Gnumbertester
Jan 29 at 22:46
$begingroup$
What do $n$ and $z$ actually represent though?
$endgroup$
– Gnumbertester
Jan 29 at 22:46
$begingroup$
n and z is fractional exponent where the index of the root just happens to equal the number we get when we make y the new exponent.
$endgroup$
– Adam Boit
Jan 30 at 5:12
$begingroup$
n and z is fractional exponent where the index of the root just happens to equal the number we get when we make y the new exponent.
$endgroup$
– Adam Boit
Jan 30 at 5:12
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Adding to abc..., note that:
$$log_b{x}=y Rightarrow b^y=x$$
You're saying:
$$2^{frac{2}{3}}approxlog_2{3}$$
If we break this into the components of a logarithm, we have:
$b=2$, $y=frac{2}{3}$, and $x=3$.
That would imply that $2^{frac{2}{3}}=3$ by our definition of a logarithm.
Clearly, this is a contradiction though as $2^{frac{2}{3}}approx1.5874$. Therefore, this is just an interesting coincidence, not consequence of any "rule".
answered Jan 29 at 21:45
GnumbertesterGnumbertester
6771114
$endgroup$
$begingroup$
Thank you a lot. I have had this sudden craze in maths and I keep bumping on some numbers now and then. I'll share more.
$endgroup$
– Adam Boit
Jan 29 at 22:48
$begingroup$
Glad I could help. I am going through the same thing, experimental math is fascinating.
$endgroup$
– Gnumbertester
Jan 29 at 23:15
add a comment |
$begingroup$
To me it's a coincidence.
$log_2 3-2^{frac{2}{3}}=0.002438551247043293297966695324491751631678920207371949352$
answered Jan 29 at 21:36
abc...abc...
3,237739
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Adding to abc..., note that:
$$log_b{x}=y Rightarrow b^y=x$$
You're saying:
$$2^{frac{2}{3}}approxlog_2{3}$$
If we break this into the components of a logarithm, we have:
$b=2$, $y=frac{2}{3}$, and $x=3$.
That would imply that $2^{frac{2}{3}}=3$ by our definition of a logarithm.
Clearly, this is a contradiction though as $2^{frac{2}{3}}approx1.5874$. Therefore, this is just an interesting coincidence, not consequence of any "rule".
answered Jan 29 at 21:45
GnumbertesterGnumbertester
6771114
$endgroup$
$begingroup$
Thank you a lot. I have had this sudden craze in maths and I keep bumping on some numbers now and then. I'll share more.
$endgroup$
– Adam Boit
Jan 29 at 22:48
$begingroup$
Glad I could help. I am going through the same thing, experimental math is fascinating.
$endgroup$
– Gnumbertester
Jan 29 at 23:15
add a comment |
$begingroup$
Adding to abc..., note that:
$$log_b{x}=y Rightarrow b^y=x$$
You're saying:
$$2^{frac{2}{3}}approxlog_2{3}$$
If we break this into the components of a logarithm, we have:
$b=2$, $y=frac{2}{3}$, and $x=3$.
That would imply that $2^{frac{2}{3}}=3$ by our definition of a logarithm.
Clearly, this is a contradiction though as $2^{frac{2}{3}}approx1.5874$. Therefore, this is just an interesting coincidence, not consequence of any "rule".
answered Jan 29 at 21:45
GnumbertesterGnumbertester
6771114
$endgroup$
$begingroup$
Thank you a lot. I have had this sudden craze in maths and I keep bumping on some numbers now and then. I'll share more.
$endgroup$
– Adam Boit
Jan 29 at 22:48
$begingroup$
Glad I could help. I am going through the same thing, experimental math is fascinating.
$endgroup$
– Gnumbertester
Jan 29 at 23:15
add a comment |
$begingroup$
Adding to abc..., note that:
$$log_b{x}=y Rightarrow b^y=x$$
You're saying:
$$2^{frac{2}{3}}approxlog_2{3}$$
If we break this into the components of a logarithm, we have:
$b=2$, $y=frac{2}{3}$, and $x=3$.
That would imply that $2^{frac{2}{3}}=3$ by our definition of a logarithm.
Clearly, this is a contradiction though as $2^{frac{2}{3}}approx1.5874$. Therefore, this is just an interesting coincidence, not consequence of any "rule".
answered Jan 29 at 21:45
GnumbertesterGnumbertester
6771114
$endgroup$
Adding to abc..., note that:
$$log_b{x}=y Rightarrow b^y=x$$
You're saying:
$$2^{frac{2}{3}}approxlog_2{3}$$
If we break this into the components of a logarithm, we have:
$b=2$, $y=frac{2}{3}$, and $x=3$.
That would imply that $2^{frac{2}{3}}=3$ by our definition of a logarithm.
Clearly, this is a contradiction though as $2^{frac{2}{3}}approx1.5874$. Therefore, this is just an interesting coincidence, not consequence of any "rule".
answered Jan 29 at 21:45
GnumbertesterGnumbertester
6771114
answered Jan 29 at 21:45
GnumbertesterGnumbertester
6771114
answered Jan 29 at 21:45
GnumbertesterGnumbertester
6771114
answered Jan 29 at 21:45
GnumbertesterGnumbertester
6771114
6771114
$begingroup$
Thank you a lot. I have had this sudden craze in maths and I keep bumping on some numbers now and then. I'll share more.
$endgroup$
– Adam Boit
Jan 29 at 22:48
$begingroup$
Glad I could help. I am going through the same thing, experimental math is fascinating.
$endgroup$
– Gnumbertester
Jan 29 at 23:15
add a comment |
$begingroup$
Thank you a lot. I have had this sudden craze in maths and I keep bumping on some numbers now and then. I'll share more.
$endgroup$
– Adam Boit
Jan 29 at 22:48
$begingroup$
Glad I could help. I am going through the same thing, experimental math is fascinating.
$endgroup$
– Gnumbertester
Jan 29 at 23:15
$begingroup$
Thank you a lot. I have had this sudden craze in maths and I keep bumping on some numbers now and then. I'll share more.
$endgroup$
– Adam Boit
Jan 29 at 22:48
$begingroup$
Thank you a lot. I have had this sudden craze in maths and I keep bumping on some numbers now and then. I'll share more.
$endgroup$
– Adam Boit
Jan 29 at 22:48
$begingroup$
Glad I could help. I am going through the same thing, experimental math is fascinating.
$endgroup$
– Gnumbertester
Jan 29 at 23:15
$begingroup$
Glad I could help. I am going through the same thing, experimental math is fascinating.
$endgroup$
– Gnumbertester
Jan 29 at 23:15
add a comment |
$begingroup$
To me it's a coincidence.
$log_2 3-2^{frac{2}{3}}=0.002438551247043293297966695324491751631678920207371949352$
answered Jan 29 at 21:36
abc...abc...
3,237739
$endgroup$
add a comment |
$begingroup$
To me it's a coincidence.
$log_2 3-2^{frac{2}{3}}=0.002438551247043293297966695324491751631678920207371949352$
answered Jan 29 at 21:36
abc...abc...
3,237739
$endgroup$
add a comment |
$begingroup$
To me it's a coincidence.
$log_2 3-2^{frac{2}{3}}=0.002438551247043293297966695324491751631678920207371949352$
answered Jan 29 at 21:36
abc...abc...
3,237739
$endgroup$
To me it's a coincidence.
$log_2 3-2^{frac{2}{3}}=0.002438551247043293297966695324491751631678920207371949352$
answered Jan 29 at 21:36
abc...abc...
3,237739
answered Jan 29 at 21:36
abc...abc...
3,237739
answered Jan 29 at 21:36
abc...abc...
3,237739
answered Jan 29 at 21:36
abc...abc...
3,237739
3,237739
add a comment |
add a comment |
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1
$begingroup$
What is n in this case?
$endgroup$
– Gnumbertester
Jan 29 at 21:30
$begingroup$
n/z = 1.99334613/3
$endgroup$
– Adam Boit
Jan 29 at 22:39
$begingroup$
What do $n$ and $z$ actually represent though?
$endgroup$
– Gnumbertester
Jan 29 at 22:46
$begingroup$
n and z is fractional exponent where the index of the root just happens to equal the number we get when we make y the new exponent.
$endgroup$
– Adam Boit
Jan 30 at 5:12