Can we simplify this logarithm? if so, please provides some tips
$begingroup$
${|x|^{11/10}} log_{|x|^{{1/10}}}|x|$.
I only know doing the first step, not sure if it is correct
$log_{|x|^{{1/10}}}(|x|^{|x|^{11/10}})$
as got stuck following this proof. Please help understand how we can get step two from step one.
logarithms
edited Jan 13 at 7:50
Siong Thye Goh
103k1468119
asked Jan 13 at 7:42
MaxfieldMaxfield
424
$endgroup$
add a comment |
$begingroup$
${|x|^{11/10}} log_{|x|^{{1/10}}}|x|$.
I only know doing the first step, not sure if it is correct
$log_{|x|^{{1/10}}}(|x|^{|x|^{11/10}})$
as got stuck following this proof. Please help understand how we can get step two from step one.
logarithms
edited Jan 13 at 7:50
Siong Thye Goh
103k1468119
asked Jan 13 at 7:42
MaxfieldMaxfield
424
$endgroup$
add a comment |
$begingroup$
${|x|^{11/10}} log_{|x|^{{1/10}}}|x|$.
I only know doing the first step, not sure if it is correct
$log_{|x|^{{1/10}}}(|x|^{|x|^{11/10}})$
as got stuck following this proof. Please help understand how we can get step two from step one.
logarithms
edited Jan 13 at 7:50
Siong Thye Goh
103k1468119
asked Jan 13 at 7:42
MaxfieldMaxfield
424
$endgroup$
${|x|^{11/10}} log_{|x|^{{1/10}}}|x|$.
I only know doing the first step, not sure if it is correct
$log_{|x|^{{1/10}}}(|x|^{|x|^{11/10}})$
as got stuck following this proof. Please help understand how we can get step two from step one.
logarithms
logarithms
edited Jan 13 at 7:50
Siong Thye Goh
103k1468119
asked Jan 13 at 7:42
MaxfieldMaxfield
424
edited Jan 13 at 7:50
Siong Thye Goh
103k1468119
asked Jan 13 at 7:42
MaxfieldMaxfield
424
edited Jan 13 at 7:50
Siong Thye Goh
103k1468119
edited Jan 13 at 7:50
Siong Thye Goh
103k1468119
edited Jan 13 at 7:50
Siong Thye Goh
103k1468119
103k1468119
asked Jan 13 at 7:42
MaxfieldMaxfield
424
asked Jan 13 at 7:42
MaxfieldMaxfield
424
asked Jan 13 at 7:42
MaxfieldMaxfield
424
424
add a comment |
add a comment |
2 Answers
2
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$begingroup$
If I interpret your question correctly, $V$ means the set of vertices and $E$ means the set of edges.
We have $|E|=O(|V|^2)$.
Hence, we can drop the first term.
Also, note that $log_ab =frac{log b}{log a}$.
answered Jan 13 at 7:51
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
add a comment |
$begingroup$
Consider
$$y=x^alog_{x^b} (x)$$ Using the laws of logarithms
$$y=frac{x^a log (x)}{log left(x^bright)}=frac{x^a log (x)}{blog left(xright)}=frac 1b x^a$$
answered Jan 13 at 10:03
Claude LeiboviciClaude Leibovici
124k1157135
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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votes
$begingroup$
If I interpret your question correctly, $V$ means the set of vertices and $E$ means the set of edges.
We have $|E|=O(|V|^2)$.
Hence, we can drop the first term.
Also, note that $log_ab =frac{log b}{log a}$.
answered Jan 13 at 7:51
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
add a comment |
$begingroup$
If I interpret your question correctly, $V$ means the set of vertices and $E$ means the set of edges.
We have $|E|=O(|V|^2)$.
Hence, we can drop the first term.
Also, note that $log_ab =frac{log b}{log a}$.
answered Jan 13 at 7:51
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
add a comment |
$begingroup$
If I interpret your question correctly, $V$ means the set of vertices and $E$ means the set of edges.
We have $|E|=O(|V|^2)$.
Hence, we can drop the first term.
Also, note that $log_ab =frac{log b}{log a}$.
answered Jan 13 at 7:51
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
If I interpret your question correctly, $V$ means the set of vertices and $E$ means the set of edges.
We have $|E|=O(|V|^2)$.
Hence, we can drop the first term.
Also, note that $log_ab =frac{log b}{log a}$.
answered Jan 13 at 7:51
Siong Thye GohSiong Thye Goh
103k1468119
answered Jan 13 at 7:51
Siong Thye GohSiong Thye Goh
103k1468119
answered Jan 13 at 7:51
Siong Thye GohSiong Thye Goh
103k1468119
answered Jan 13 at 7:51
Siong Thye GohSiong Thye Goh
103k1468119
103k1468119
add a comment |
add a comment |
$begingroup$
Consider
$$y=x^alog_{x^b} (x)$$ Using the laws of logarithms
$$y=frac{x^a log (x)}{log left(x^bright)}=frac{x^a log (x)}{blog left(xright)}=frac 1b x^a$$
answered Jan 13 at 10:03
Claude LeiboviciClaude Leibovici
124k1157135
$endgroup$
add a comment |
$begingroup$
Consider
$$y=x^alog_{x^b} (x)$$ Using the laws of logarithms
$$y=frac{x^a log (x)}{log left(x^bright)}=frac{x^a log (x)}{blog left(xright)}=frac 1b x^a$$
answered Jan 13 at 10:03
Claude LeiboviciClaude Leibovici
124k1157135
$endgroup$
add a comment |
$begingroup$
Consider
$$y=x^alog_{x^b} (x)$$ Using the laws of logarithms
$$y=frac{x^a log (x)}{log left(x^bright)}=frac{x^a log (x)}{blog left(xright)}=frac 1b x^a$$
answered Jan 13 at 10:03
Claude LeiboviciClaude Leibovici
124k1157135
$endgroup$
Consider
$$y=x^alog_{x^b} (x)$$ Using the laws of logarithms
$$y=frac{x^a log (x)}{log left(x^bright)}=frac{x^a log (x)}{blog left(xright)}=frac 1b x^a$$
answered Jan 13 at 10:03
Claude LeiboviciClaude Leibovici
124k1157135
answered Jan 13 at 10:03
Claude LeiboviciClaude Leibovici
124k1157135
answered Jan 13 at 10:03
Claude LeiboviciClaude Leibovici
124k1157135
answered Jan 13 at 10:03
Claude LeiboviciClaude Leibovici
124k1157135
124k1157135
add a comment |
add a comment |
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