Mixing
$x^{log_5 3}=-6$,Solve for x
$begingroup$
This question feels banal to ask, but I feel lost. Here is what I did.
$x^{log_5 3}=-6$
$ln(x^{log_5 3})=ln(-6)=ln(6)+ipi$
$log_53* ln(x)=ln(6)+ipi$
$ln(x)=frac{ln(6)+ipi}{log_53}=(ln(6)+ipi)log_35$
$x=e^{ln(6)+ipi)log_35}$
This seems right to me, but when I verify with wolfram it shows that I am wrong.
https://www.wolframalpha.com/input/?i=(e%5E(ln(-6)log(3,5)))%5E(log(5,3))
proof-verification logarithms
edited Jan 31 at 4:55
YuiTo Cheng
2,2634937
asked Jan 31 at 3:10
Alex HalAlex Hal
537
$endgroup$
add a comment |
$begingroup$
This question feels banal to ask, but I feel lost. Here is what I did.
$x^{log_5 3}=-6$
$ln(x^{log_5 3})=ln(-6)=ln(6)+ipi$
$log_53* ln(x)=ln(6)+ipi$
$ln(x)=frac{ln(6)+ipi}{log_53}=(ln(6)+ipi)log_35$
$x=e^{ln(6)+ipi)log_35}$
This seems right to me, but when I verify with wolfram it shows that I am wrong.
https://www.wolframalpha.com/input/?i=(e%5E(ln(-6)log(3,5)))%5E(log(5,3))
proof-verification logarithms
edited Jan 31 at 4:55
YuiTo Cheng
2,2634937
asked Jan 31 at 3:10
Alex HalAlex Hal
537
$endgroup$
2
$begingroup$
Putting aside my nervousness at using ln when it’s not well-defined, I think your computation looks right, too, but the result doesn’t. I’d expect your answer to look like $e^{text{(real) }+ipi}$.
$endgroup$
– Lubin
Jan 31 at 3:20
$begingroup$
Yeah, the line of reasoning seems to be fine.
$endgroup$
– s0ulr3aper07
Jan 31 at 3:28
$begingroup$
I do agree, the result doesn't look like the ones I usually see. But it seems to work with the computation. I'm starting to believe that wolfram made an error but I think I made one somewhere.
$endgroup$
– Alex Hal
Jan 31 at 3:37
$begingroup$
As my updated answer shows, I originally made a mistake. The issue with your use of Wolfram is not double exponentiation but, instead, the extra use of brackets causing it to interpret things differently. I'm sorry for this error, so please reread my answer for the appropriate details.
$endgroup$
– John Omielan
Jan 31 at 13:08
add a comment |
$begingroup$
This question feels banal to ask, but I feel lost. Here is what I did.
$x^{log_5 3}=-6$
$ln(x^{log_5 3})=ln(-6)=ln(6)+ipi$
$log_53* ln(x)=ln(6)+ipi$
$ln(x)=frac{ln(6)+ipi}{log_53}=(ln(6)+ipi)log_35$
$x=e^{ln(6)+ipi)log_35}$
This seems right to me, but when I verify with wolfram it shows that I am wrong.
https://www.wolframalpha.com/input/?i=(e%5E(ln(-6)log(3,5)))%5E(log(5,3))
proof-verification logarithms
edited Jan 31 at 4:55
YuiTo Cheng
2,2634937
asked Jan 31 at 3:10
Alex HalAlex Hal
537
$endgroup$
This question feels banal to ask, but I feel lost. Here is what I did.
$x^{log_5 3}=-6$
$ln(x^{log_5 3})=ln(-6)=ln(6)+ipi$
$log_53* ln(x)=ln(6)+ipi$
$ln(x)=frac{ln(6)+ipi}{log_53}=(ln(6)+ipi)log_35$
$x=e^{ln(6)+ipi)log_35}$
This seems right to me, but when I verify with wolfram it shows that I am wrong.
https://www.wolframalpha.com/input/?i=(e%5E(ln(-6)log(3,5)))%5E(log(5,3))
proof-verification logarithms
proof-verification logarithms
edited Jan 31 at 4:55
YuiTo Cheng
2,2634937
asked Jan 31 at 3:10
Alex HalAlex Hal
537
edited Jan 31 at 4:55
YuiTo Cheng
2,2634937
asked Jan 31 at 3:10
Alex HalAlex Hal
537
edited Jan 31 at 4:55
YuiTo Cheng
2,2634937
edited Jan 31 at 4:55
YuiTo Cheng
2,2634937
edited Jan 31 at 4:55
YuiTo Cheng
2,2634937
2,2634937
asked Jan 31 at 3:10
Alex HalAlex Hal
537
asked Jan 31 at 3:10
Alex HalAlex Hal
537
asked Jan 31 at 3:10
Alex HalAlex Hal
537
537
2
$begingroup$
Putting aside my nervousness at using ln when it’s not well-defined, I think your computation looks right, too, but the result doesn’t. I’d expect your answer to look like $e^{text{(real) }+ipi}$.
$endgroup$
– Lubin
Jan 31 at 3:20
$begingroup$
Yeah, the line of reasoning seems to be fine.
$endgroup$
– s0ulr3aper07
Jan 31 at 3:28
$begingroup$
I do agree, the result doesn't look like the ones I usually see. But it seems to work with the computation. I'm starting to believe that wolfram made an error but I think I made one somewhere.
$endgroup$
– Alex Hal
Jan 31 at 3:37
$begingroup$
As my updated answer shows, I originally made a mistake. The issue with your use of Wolfram is not double exponentiation but, instead, the extra use of brackets causing it to interpret things differently. I'm sorry for this error, so please reread my answer for the appropriate details.
$endgroup$
– John Omielan
Jan 31 at 13:08
add a comment |
2
$begingroup$
Putting aside my nervousness at using ln when it’s not well-defined, I think your computation looks right, too, but the result doesn’t. I’d expect your answer to look like $e^{text{(real) }+ipi}$.
$endgroup$
– Lubin
Jan 31 at 3:20
$begingroup$
Yeah, the line of reasoning seems to be fine.
$endgroup$
– s0ulr3aper07
Jan 31 at 3:28
$begingroup$
I do agree, the result doesn't look like the ones I usually see. But it seems to work with the computation. I'm starting to believe that wolfram made an error but I think I made one somewhere.
$endgroup$
– Alex Hal
Jan 31 at 3:37
$begingroup$
As my updated answer shows, I originally made a mistake. The issue with your use of Wolfram is not double exponentiation but, instead, the extra use of brackets causing it to interpret things differently. I'm sorry for this error, so please reread my answer for the appropriate details.
$endgroup$
– John Omielan
Jan 31 at 13:08
2
2
$begingroup$
Putting aside my nervousness at using ln when it’s not well-defined, I think your computation looks right, too, but the result doesn’t. I’d expect your answer to look like $e^{text{(real) }+ipi}$.
$endgroup$
– Lubin
Jan 31 at 3:20
$begingroup$
Putting aside my nervousness at using ln when it’s not well-defined, I think your computation looks right, too, but the result doesn’t. I’d expect your answer to look like $e^{text{(real) }+ipi}$.
$endgroup$
– Lubin
Jan 31 at 3:20
$begingroup$
Yeah, the line of reasoning seems to be fine.
$endgroup$
– s0ulr3aper07
Jan 31 at 3:28
$begingroup$
Yeah, the line of reasoning seems to be fine.
$endgroup$
– s0ulr3aper07
Jan 31 at 3:28
$begingroup$
I do agree, the result doesn't look like the ones I usually see. But it seems to work with the computation. I'm starting to believe that wolfram made an error but I think I made one somewhere.
$endgroup$
– Alex Hal
Jan 31 at 3:37
$begingroup$
I do agree, the result doesn't look like the ones I usually see. But it seems to work with the computation. I'm starting to believe that wolfram made an error but I think I made one somewhere.
$endgroup$
– Alex Hal
Jan 31 at 3:37
$begingroup$
As my updated answer shows, I originally made a mistake. The issue with your use of Wolfram is not double exponentiation but, instead, the extra use of brackets causing it to interpret things differently. I'm sorry for this error, so please reread my answer for the appropriate details.
$endgroup$
– John Omielan
Jan 31 at 13:08
$begingroup$
As my updated answer shows, I originally made a mistake. The issue with your use of Wolfram is not double exponentiation but, instead, the extra use of brackets causing it to interpret things differently. I'm sorry for this error, so please reread my answer for the appropriate details.
$endgroup$
– John Omielan
Jan 31 at 13:08
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Update: I originally thought the input text to Wolfram was causing a double exponentiation. I later realized this is not true, with the issue instead being due to how the use of brackets caused it to interpret the calculations. I have now corrected my answer below.
$ $
As commented by several people, your solution is basically correct. The issue with your use of Wolfram is that your input text is causing it to interpret it differently than you intended due to your use of brackets. You asked it to find "(e^(ln(-6)log(3,5)))^(log(5,3))". Using your specific instructions, Wolfram is calculating, as it says in the exact result section:
$$left(5^{left(left(log(6) + i πright)/logleft(3right)right)}right)^{(log(3)/log(5))} tag{1}label{eq1}$$
i.e., it's using a base of $5$ instead of the intended $6$. This is due to your use of brackets and a power separate from the original expression, causing Wolfram to reinterpret the part inside the large brackets on their own, where the radius in polar coordinates is $5$, before then applying the final power. However, as the "Polar coordinates" part shows, the radius of the overall expression is still $6$ (as the exponent outside of the main brackets of $logleft(3right)/logleft(5right)$ causes it to be changed), but with a $theta$ of about $-65.7382°$ instead of the $pi$ radians, i.e., $180°$, which you wanted. As Mark Viola says in his answer, the logarithm function with complex variables is multi-valued. Thus, you need to be careful how you use it, including how you express it to Wolfram to calculate for you.
The better expression to use with Wolfram is something like "e^((ln(-6)log(3,5))(log(5,3)))" (note: I'm not using a direct link with the equation in Wolfram as the URL is not being handled properly), i.e., remove your second "^" and adjust your braces. This puts all of your work together within the same brackets, in particular it has the multiplication of the exponents together, so Wolfram interprets it correctly for you, i.e., it gives you back a result of $-6$ as you expected.
answered Jan 31 at 5:36
John OmielanJohn Omielan
4,6312215
$endgroup$
add a comment |
$begingroup$
The definition of $z^C$, where $zinmathbb{C}$ and $Cin mathbb{C}$, is the multivalued number given by
$$z^C=e^{Cleft(log(|z|+iarg(z)) right)}$$
where $arg(z)$ is multivalued and $log(|z|)$ is the logarithm from real analysis.
Now, suppose $C=log_5(3)$ and $z=xinmathbb{C}$, we have
$$x^{log_5(3)}=|x|^{log_5(3)}e^{ilog_5(3)arg(x)}$$
Setting $x^{log_5(3)}=-6=6e^{i(2n+1)pi}$ we find that the magnitude of $x$ is
$$|x|=6^{log(5)/log(3)}$$
and the argument of $x$ is
$$arg(x)=(2n+1)pifrac{log(5)}{log(3)}$$
Therefore, the solution for $x$ is
$$x=e^{log(5)log(6)/log(3)}e^{i(2n+1)pifrac{log(5)}{log(3)}}$$
answered Jan 31 at 5:14
Mark ViolaMark Viola
134k1278177
$endgroup$
$begingroup$
This also is false according to wolfram... wolframalpha.com/input/…
$endgroup$
– Alex Hal
Jan 31 at 5:28
$begingroup$
@AlexHal Actually, if you look at the last "box" in your WA link, you will see that when $n$ is a real valued number, the expression in my solution raised to the $log_5(3)$ is indeed $-6$. It reads "Alternate form assuming n is real:"
$endgroup$
– Mark Viola
Jan 31 at 5:41
$begingroup$
I posted the correct solution, which DOES agree with WA.
$endgroup$
– Mark Viola
Jan 31 at 6:05
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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$begingroup$
Update: I originally thought the input text to Wolfram was causing a double exponentiation. I later realized this is not true, with the issue instead being due to how the use of brackets caused it to interpret the calculations. I have now corrected my answer below.
$ $
As commented by several people, your solution is basically correct. The issue with your use of Wolfram is that your input text is causing it to interpret it differently than you intended due to your use of brackets. You asked it to find "(e^(ln(-6)log(3,5)))^(log(5,3))". Using your specific instructions, Wolfram is calculating, as it says in the exact result section:
$$left(5^{left(left(log(6) + i πright)/logleft(3right)right)}right)^{(log(3)/log(5))} tag{1}label{eq1}$$
i.e., it's using a base of $5$ instead of the intended $6$. This is due to your use of brackets and a power separate from the original expression, causing Wolfram to reinterpret the part inside the large brackets on their own, where the radius in polar coordinates is $5$, before then applying the final power. However, as the "Polar coordinates" part shows, the radius of the overall expression is still $6$ (as the exponent outside of the main brackets of $logleft(3right)/logleft(5right)$ causes it to be changed), but with a $theta$ of about $-65.7382°$ instead of the $pi$ radians, i.e., $180°$, which you wanted. As Mark Viola says in his answer, the logarithm function with complex variables is multi-valued. Thus, you need to be careful how you use it, including how you express it to Wolfram to calculate for you.
The better expression to use with Wolfram is something like "e^((ln(-6)log(3,5))(log(5,3)))" (note: I'm not using a direct link with the equation in Wolfram as the URL is not being handled properly), i.e., remove your second "^" and adjust your braces. This puts all of your work together within the same brackets, in particular it has the multiplication of the exponents together, so Wolfram interprets it correctly for you, i.e., it gives you back a result of $-6$ as you expected.
answered Jan 31 at 5:36
John OmielanJohn Omielan
4,6312215
$endgroup$
add a comment |
$begingroup$
Update: I originally thought the input text to Wolfram was causing a double exponentiation. I later realized this is not true, with the issue instead being due to how the use of brackets caused it to interpret the calculations. I have now corrected my answer below.
$ $
As commented by several people, your solution is basically correct. The issue with your use of Wolfram is that your input text is causing it to interpret it differently than you intended due to your use of brackets. You asked it to find "(e^(ln(-6)log(3,5)))^(log(5,3))". Using your specific instructions, Wolfram is calculating, as it says in the exact result section:
$$left(5^{left(left(log(6) + i πright)/logleft(3right)right)}right)^{(log(3)/log(5))} tag{1}label{eq1}$$
i.e., it's using a base of $5$ instead of the intended $6$. This is due to your use of brackets and a power separate from the original expression, causing Wolfram to reinterpret the part inside the large brackets on their own, where the radius in polar coordinates is $5$, before then applying the final power. However, as the "Polar coordinates" part shows, the radius of the overall expression is still $6$ (as the exponent outside of the main brackets of $logleft(3right)/logleft(5right)$ causes it to be changed), but with a $theta$ of about $-65.7382°$ instead of the $pi$ radians, i.e., $180°$, which you wanted. As Mark Viola says in his answer, the logarithm function with complex variables is multi-valued. Thus, you need to be careful how you use it, including how you express it to Wolfram to calculate for you.
The better expression to use with Wolfram is something like "e^((ln(-6)log(3,5))(log(5,3)))" (note: I'm not using a direct link with the equation in Wolfram as the URL is not being handled properly), i.e., remove your second "^" and adjust your braces. This puts all of your work together within the same brackets, in particular it has the multiplication of the exponents together, so Wolfram interprets it correctly for you, i.e., it gives you back a result of $-6$ as you expected.
answered Jan 31 at 5:36
John OmielanJohn Omielan
4,6312215
$endgroup$
add a comment |
$begingroup$
Update: I originally thought the input text to Wolfram was causing a double exponentiation. I later realized this is not true, with the issue instead being due to how the use of brackets caused it to interpret the calculations. I have now corrected my answer below.
$ $
As commented by several people, your solution is basically correct. The issue with your use of Wolfram is that your input text is causing it to interpret it differently than you intended due to your use of brackets. You asked it to find "(e^(ln(-6)log(3,5)))^(log(5,3))". Using your specific instructions, Wolfram is calculating, as it says in the exact result section:
$$left(5^{left(left(log(6) + i πright)/logleft(3right)right)}right)^{(log(3)/log(5))} tag{1}label{eq1}$$
i.e., it's using a base of $5$ instead of the intended $6$. This is due to your use of brackets and a power separate from the original expression, causing Wolfram to reinterpret the part inside the large brackets on their own, where the radius in polar coordinates is $5$, before then applying the final power. However, as the "Polar coordinates" part shows, the radius of the overall expression is still $6$ (as the exponent outside of the main brackets of $logleft(3right)/logleft(5right)$ causes it to be changed), but with a $theta$ of about $-65.7382°$ instead of the $pi$ radians, i.e., $180°$, which you wanted. As Mark Viola says in his answer, the logarithm function with complex variables is multi-valued. Thus, you need to be careful how you use it, including how you express it to Wolfram to calculate for you.
The better expression to use with Wolfram is something like "e^((ln(-6)log(3,5))(log(5,3)))" (note: I'm not using a direct link with the equation in Wolfram as the URL is not being handled properly), i.e., remove your second "^" and adjust your braces. This puts all of your work together within the same brackets, in particular it has the multiplication of the exponents together, so Wolfram interprets it correctly for you, i.e., it gives you back a result of $-6$ as you expected.
answered Jan 31 at 5:36
John OmielanJohn Omielan
4,6312215
$endgroup$
Update: I originally thought the input text to Wolfram was causing a double exponentiation. I later realized this is not true, with the issue instead being due to how the use of brackets caused it to interpret the calculations. I have now corrected my answer below.
$ $
As commented by several people, your solution is basically correct. The issue with your use of Wolfram is that your input text is causing it to interpret it differently than you intended due to your use of brackets. You asked it to find "(e^(ln(-6)log(3,5)))^(log(5,3))". Using your specific instructions, Wolfram is calculating, as it says in the exact result section:
$$left(5^{left(left(log(6) + i πright)/logleft(3right)right)}right)^{(log(3)/log(5))} tag{1}label{eq1}$$
i.e., it's using a base of $5$ instead of the intended $6$. This is due to your use of brackets and a power separate from the original expression, causing Wolfram to reinterpret the part inside the large brackets on their own, where the radius in polar coordinates is $5$, before then applying the final power. However, as the "Polar coordinates" part shows, the radius of the overall expression is still $6$ (as the exponent outside of the main brackets of $logleft(3right)/logleft(5right)$ causes it to be changed), but with a $theta$ of about $-65.7382°$ instead of the $pi$ radians, i.e., $180°$, which you wanted. As Mark Viola says in his answer, the logarithm function with complex variables is multi-valued. Thus, you need to be careful how you use it, including how you express it to Wolfram to calculate for you.
The better expression to use with Wolfram is something like "e^((ln(-6)log(3,5))(log(5,3)))" (note: I'm not using a direct link with the equation in Wolfram as the URL is not being handled properly), i.e., remove your second "^" and adjust your braces. This puts all of your work together within the same brackets, in particular it has the multiplication of the exponents together, so Wolfram interprets it correctly for you, i.e., it gives you back a result of $-6$ as you expected.
answered Jan 31 at 5:36
John OmielanJohn Omielan
4,6312215
edited Jan 31 at 13:04
answered Jan 31 at 5:36
John OmielanJohn Omielan
4,6312215
answered Jan 31 at 5:36
John OmielanJohn Omielan
4,6312215
answered Jan 31 at 5:36
John OmielanJohn Omielan
4,6312215
4,6312215
add a comment |
add a comment |
$begingroup$
The definition of $z^C$, where $zinmathbb{C}$ and $Cin mathbb{C}$, is the multivalued number given by
$$z^C=e^{Cleft(log(|z|+iarg(z)) right)}$$
where $arg(z)$ is multivalued and $log(|z|)$ is the logarithm from real analysis.
Now, suppose $C=log_5(3)$ and $z=xinmathbb{C}$, we have
$$x^{log_5(3)}=|x|^{log_5(3)}e^{ilog_5(3)arg(x)}$$
Setting $x^{log_5(3)}=-6=6e^{i(2n+1)pi}$ we find that the magnitude of $x$ is
$$|x|=6^{log(5)/log(3)}$$
and the argument of $x$ is
$$arg(x)=(2n+1)pifrac{log(5)}{log(3)}$$
Therefore, the solution for $x$ is
$$x=e^{log(5)log(6)/log(3)}e^{i(2n+1)pifrac{log(5)}{log(3)}}$$
answered Jan 31 at 5:14
Mark ViolaMark Viola
134k1278177
$endgroup$
$begingroup$
This also is false according to wolfram... wolframalpha.com/input/…
$endgroup$
– Alex Hal
Jan 31 at 5:28
$begingroup$
@AlexHal Actually, if you look at the last "box" in your WA link, you will see that when $n$ is a real valued number, the expression in my solution raised to the $log_5(3)$ is indeed $-6$. It reads "Alternate form assuming n is real:"
$endgroup$
– Mark Viola
Jan 31 at 5:41
$begingroup$
I posted the correct solution, which DOES agree with WA.
$endgroup$
– Mark Viola
Jan 31 at 6:05
add a comment |
$begingroup$
The definition of $z^C$, where $zinmathbb{C}$ and $Cin mathbb{C}$, is the multivalued number given by
$$z^C=e^{Cleft(log(|z|+iarg(z)) right)}$$
where $arg(z)$ is multivalued and $log(|z|)$ is the logarithm from real analysis.
Now, suppose $C=log_5(3)$ and $z=xinmathbb{C}$, we have
$$x^{log_5(3)}=|x|^{log_5(3)}e^{ilog_5(3)arg(x)}$$
Setting $x^{log_5(3)}=-6=6e^{i(2n+1)pi}$ we find that the magnitude of $x$ is
$$|x|=6^{log(5)/log(3)}$$
and the argument of $x$ is
$$arg(x)=(2n+1)pifrac{log(5)}{log(3)}$$
Therefore, the solution for $x$ is
$$x=e^{log(5)log(6)/log(3)}e^{i(2n+1)pifrac{log(5)}{log(3)}}$$
answered Jan 31 at 5:14
Mark ViolaMark Viola
134k1278177
$endgroup$
$begingroup$
This also is false according to wolfram... wolframalpha.com/input/…
$endgroup$
– Alex Hal
Jan 31 at 5:28
$begingroup$
@AlexHal Actually, if you look at the last "box" in your WA link, you will see that when $n$ is a real valued number, the expression in my solution raised to the $log_5(3)$ is indeed $-6$. It reads "Alternate form assuming n is real:"
$endgroup$
– Mark Viola
Jan 31 at 5:41
$begingroup$
I posted the correct solution, which DOES agree with WA.
$endgroup$
– Mark Viola
Jan 31 at 6:05
add a comment |
$begingroup$
The definition of $z^C$, where $zinmathbb{C}$ and $Cin mathbb{C}$, is the multivalued number given by
$$z^C=e^{Cleft(log(|z|+iarg(z)) right)}$$
where $arg(z)$ is multivalued and $log(|z|)$ is the logarithm from real analysis.
Now, suppose $C=log_5(3)$ and $z=xinmathbb{C}$, we have
$$x^{log_5(3)}=|x|^{log_5(3)}e^{ilog_5(3)arg(x)}$$
Setting $x^{log_5(3)}=-6=6e^{i(2n+1)pi}$ we find that the magnitude of $x$ is
$$|x|=6^{log(5)/log(3)}$$
and the argument of $x$ is
$$arg(x)=(2n+1)pifrac{log(5)}{log(3)}$$
Therefore, the solution for $x$ is
$$x=e^{log(5)log(6)/log(3)}e^{i(2n+1)pifrac{log(5)}{log(3)}}$$
answered Jan 31 at 5:14
Mark ViolaMark Viola
134k1278177
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The definition of $z^C$, where $zinmathbb{C}$ and $Cin mathbb{C}$, is the multivalued number given by
$$z^C=e^{Cleft(log(|z|+iarg(z)) right)}$$
where $arg(z)$ is multivalued and $log(|z|)$ is the logarithm from real analysis.
Now, suppose $C=log_5(3)$ and $z=xinmathbb{C}$, we have
$$x^{log_5(3)}=|x|^{log_5(3)}e^{ilog_5(3)arg(x)}$$
Setting $x^{log_5(3)}=-6=6e^{i(2n+1)pi}$ we find that the magnitude of $x$ is
$$|x|=6^{log(5)/log(3)}$$
and the argument of $x$ is
$$arg(x)=(2n+1)pifrac{log(5)}{log(3)}$$
Therefore, the solution for $x$ is
$$x=e^{log(5)log(6)/log(3)}e^{i(2n+1)pifrac{log(5)}{log(3)}}$$
answered Jan 31 at 5:14
Mark ViolaMark Viola
134k1278177
answered Jan 31 at 5:14
Mark ViolaMark Viola
134k1278177
answered Jan 31 at 5:14
Mark ViolaMark Viola
134k1278177
answered Jan 31 at 5:14
Mark ViolaMark Viola
134k1278177
134k1278177
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This also is false according to wolfram... wolframalpha.com/input/…
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– Alex Hal
Jan 31 at 5:28
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@AlexHal Actually, if you look at the last "box" in your WA link, you will see that when $n$ is a real valued number, the expression in my solution raised to the $log_5(3)$ is indeed $-6$. It reads "Alternate form assuming n is real:"
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– Mark Viola
Jan 31 at 5:41
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I posted the correct solution, which DOES agree with WA.
$endgroup$
– Mark Viola
Jan 31 at 6:05
add a comment |
$begingroup$
This also is false according to wolfram... wolframalpha.com/input/…
$endgroup$
– Alex Hal
Jan 31 at 5:28
$begingroup$
@AlexHal Actually, if you look at the last "box" in your WA link, you will see that when $n$ is a real valued number, the expression in my solution raised to the $log_5(3)$ is indeed $-6$. It reads "Alternate form assuming n is real:"
$endgroup$
– Mark Viola
Jan 31 at 5:41
$begingroup$
I posted the correct solution, which DOES agree with WA.
$endgroup$
– Mark Viola
Jan 31 at 6:05
$begingroup$
This also is false according to wolfram... wolframalpha.com/input/…
$endgroup$
– Alex Hal
Jan 31 at 5:28
$begingroup$
This also is false according to wolfram... wolframalpha.com/input/…
$endgroup$
– Alex Hal
Jan 31 at 5:28
$begingroup$
@AlexHal Actually, if you look at the last "box" in your WA link, you will see that when $n$ is a real valued number, the expression in my solution raised to the $log_5(3)$ is indeed $-6$. It reads "Alternate form assuming n is real:"
$endgroup$
– Mark Viola
Jan 31 at 5:41
$begingroup$
@AlexHal Actually, if you look at the last "box" in your WA link, you will see that when $n$ is a real valued number, the expression in my solution raised to the $log_5(3)$ is indeed $-6$. It reads "Alternate form assuming n is real:"
$endgroup$
– Mark Viola
Jan 31 at 5:41
$begingroup$
I posted the correct solution, which DOES agree with WA.
$endgroup$
– Mark Viola
Jan 31 at 6:05
$begingroup$
I posted the correct solution, which DOES agree with WA.
$endgroup$
– Mark Viola
Jan 31 at 6:05
add a comment |
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Putting aside my nervousness at using ln when it’s not well-defined, I think your computation looks right, too, but the result doesn’t. I’d expect your answer to look like $e^{text{(real) }+ipi}$.
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– Lubin
Jan 31 at 3:20
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Yeah, the line of reasoning seems to be fine.
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– s0ulr3aper07
Jan 31 at 3:28
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I do agree, the result doesn't look like the ones I usually see. But it seems to work with the computation. I'm starting to believe that wolfram made an error but I think I made one somewhere.
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– Alex Hal
Jan 31 at 3:37
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As my updated answer shows, I originally made a mistake. The issue with your use of Wolfram is not double exponentiation but, instead, the extra use of brackets causing it to interpret things differently. I'm sorry for this error, so please reread my answer for the appropriate details.
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– John Omielan
Jan 31 at 13:08