Find $n$ in a log equation












1












$begingroup$


I am having trouble solving this problem.




Suppose we are comparing implementations of insertion sort and merge sort on the same machine. For inputs of size $n$, insertion sort runs in $8n^2$ steps, while merge sort runs in $64 nlog n$ steps. For which values of $n$ does insertion sort beat merge sort?




*$log$ is log base $10$



$rightarrow 8n^2 = 64nlog n$ Divide both sides by $n$
$rightarrow 8n = 64log n$ Divide both sides by $64$
$rightarrow frac{1}{8} = log n$
$rightarrow log_{10}n = frac{1}{8}$
$rightarrow n = 10^{frac{1}{8}}$



When I plug $10^{frac{1}{8}}$ in as $n$, I get $14.215 = 10.649$ so this doesn't seem to add up. Can someone help me understand what I am doing wrong?






logarithms







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    You divided by 64 not by 64n.
    $endgroup$
    – hamam_Abdallah
    Dec 17 '18 at 20:01










  • $begingroup$
    I can't divide by 64n because I canceled n in 64n out in the first step
    $endgroup$
    – Evan Kim
    Dec 17 '18 at 20:05






  • 2




    $begingroup$
    @EvanKim $frac n8=log n$
    $endgroup$
    – Shubham Johri
    Dec 17 '18 at 20:11












  • $begingroup$
    Oh I see it now
    $endgroup$
    – Evan Kim
    Dec 17 '18 at 20:13
















1












$begingroup$


I am having trouble solving this problem.




Suppose we are comparing implementations of insertion sort and merge sort on the same machine. For inputs of size $n$, insertion sort runs in $8n^2$ steps, while merge sort runs in $64 nlog n$ steps. For which values of $n$ does insertion sort beat merge sort?




*$log$ is log base $10$



$rightarrow 8n^2 = 64nlog n$ Divide both sides by $n$
$rightarrow 8n = 64log n$ Divide both sides by $64$
$rightarrow frac{1}{8} = log n$
$rightarrow log_{10}n = frac{1}{8}$
$rightarrow n = 10^{frac{1}{8}}$



When I plug $10^{frac{1}{8}}$ in as $n$, I get $14.215 = 10.649$ so this doesn't seem to add up. Can someone help me understand what I am doing wrong?






logarithms







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    You divided by 64 not by 64n.
    $endgroup$
    – hamam_Abdallah
    Dec 17 '18 at 20:01










  • $begingroup$
    I can't divide by 64n because I canceled n in 64n out in the first step
    $endgroup$
    – Evan Kim
    Dec 17 '18 at 20:05






  • 2




    $begingroup$
    @EvanKim $frac n8=log n$
    $endgroup$
    – Shubham Johri
    Dec 17 '18 at 20:11












  • $begingroup$
    Oh I see it now
    $endgroup$
    – Evan Kim
    Dec 17 '18 at 20:13














1












1








1





$begingroup$


I am having trouble solving this problem.




Suppose we are comparing implementations of insertion sort and merge sort on the same machine. For inputs of size $n$, insertion sort runs in $8n^2$ steps, while merge sort runs in $64 nlog n$ steps. For which values of $n$ does insertion sort beat merge sort?




*$log$ is log base $10$



$rightarrow 8n^2 = 64nlog n$ Divide both sides by $n$
$rightarrow 8n = 64log n$ Divide both sides by $64$
$rightarrow frac{1}{8} = log n$
$rightarrow log_{10}n = frac{1}{8}$
$rightarrow n = 10^{frac{1}{8}}$



When I plug $10^{frac{1}{8}}$ in as $n$, I get $14.215 = 10.649$ so this doesn't seem to add up. Can someone help me understand what I am doing wrong?






logarithms







share|cite|improve this question











$endgroup$




I am having trouble solving this problem.




Suppose we are comparing implementations of insertion sort and merge sort on the same machine. For inputs of size $n$, insertion sort runs in $8n^2$ steps, while merge sort runs in $64 nlog n$ steps. For which values of $n$ does insertion sort beat merge sort?




*$log$ is log base $10$



$rightarrow 8n^2 = 64nlog n$ Divide both sides by $n$
$rightarrow 8n = 64log n$ Divide both sides by $64$
$rightarrow frac{1}{8} = log n$
$rightarrow log_{10}n = frac{1}{8}$
$rightarrow n = 10^{frac{1}{8}}$



When I plug $10^{frac{1}{8}}$ in as $n$, I get $14.215 = 10.649$ so this doesn't seem to add up. Can someone help me understand what I am doing wrong?






logarithms




logarithms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 21:15









Yadati Kiran

1,7911619




1,7911619










asked Dec 17 '18 at 19:58









Evan KimEvan Kim

3568




3568








  • 2




    $begingroup$
    You divided by 64 not by 64n.
    $endgroup$
    – hamam_Abdallah
    Dec 17 '18 at 20:01










  • $begingroup$
    I can't divide by 64n because I canceled n in 64n out in the first step
    $endgroup$
    – Evan Kim
    Dec 17 '18 at 20:05






  • 2




    $begingroup$
    @EvanKim $frac n8=log n$
    $endgroup$
    – Shubham Johri
    Dec 17 '18 at 20:11












  • $begingroup$
    Oh I see it now
    $endgroup$
    – Evan Kim
    Dec 17 '18 at 20:13














  • 2




    $begingroup$
    You divided by 64 not by 64n.
    $endgroup$
    – hamam_Abdallah
    Dec 17 '18 at 20:01










  • $begingroup$
    I can't divide by 64n because I canceled n in 64n out in the first step
    $endgroup$
    – Evan Kim
    Dec 17 '18 at 20:05






  • 2




    $begingroup$
    @EvanKim $frac n8=log n$
    $endgroup$
    – Shubham Johri
    Dec 17 '18 at 20:11












  • $begingroup$
    Oh I see it now
    $endgroup$
    – Evan Kim
    Dec 17 '18 at 20:13








2




2




$begingroup$
You divided by 64 not by 64n.
$endgroup$
– hamam_Abdallah
Dec 17 '18 at 20:01




$begingroup$
You divided by 64 not by 64n.
$endgroup$
– hamam_Abdallah
Dec 17 '18 at 20:01












$begingroup$
I can't divide by 64n because I canceled n in 64n out in the first step
$endgroup$
– Evan Kim
Dec 17 '18 at 20:05




$begingroup$
I can't divide by 64n because I canceled n in 64n out in the first step
$endgroup$
– Evan Kim
Dec 17 '18 at 20:05




2




2




$begingroup$
@EvanKim $frac n8=log n$
$endgroup$
– Shubham Johri
Dec 17 '18 at 20:11






$begingroup$
@EvanKim $frac n8=log n$
$endgroup$
– Shubham Johri
Dec 17 '18 at 20:11














$begingroup$
Oh I see it now
$endgroup$
– Evan Kim
Dec 17 '18 at 20:13




$begingroup$
Oh I see it now
$endgroup$
– Evan Kim
Dec 17 '18 at 20:13










1 Answer
1






active

oldest

votes


















1












$begingroup$

You have a wrong simplification: the inequality $8n^2<64nlog n$ becomes
$$
n<8log n
$$
You can check that the function $f(x)=8log x-x$ has a maximum at $8log eapprox 3.47$, and so it is decreasing for $x>8log e$. Thus when you have found $nge 4$ satisfying the inequality, with $n+1$ not satisfying it, you're done.



We have $8log 6approx6.22>6$ and $8log 7approx6.76<7$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Forgive me for my question, but why did you use log$e$?
    $endgroup$
    – Evan Kim
    Dec 18 '18 at 0:45












  • $begingroup$
    @EvanKim Because the logarithm is in base 10.
    $endgroup$
    – egreg
    Dec 18 '18 at 7:44













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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

oldest

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active

oldest

votes









1












$begingroup$

You have a wrong simplification: the inequality $8n^2<64nlog n$ becomes
$$
n<8log n
$$
You can check that the function $f(x)=8log x-x$ has a maximum at $8log eapprox 3.47$, and so it is decreasing for $x>8log e$. Thus when you have found $nge 4$ satisfying the inequality, with $n+1$ not satisfying it, you're done.



We have $8log 6approx6.22>6$ and $8log 7approx6.76<7$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Forgive me for my question, but why did you use log$e$?
    $endgroup$
    – Evan Kim
    Dec 18 '18 at 0:45












  • $begingroup$
    @EvanKim Because the logarithm is in base 10.
    $endgroup$
    – egreg
    Dec 18 '18 at 7:44


















1












$begingroup$

You have a wrong simplification: the inequality $8n^2<64nlog n$ becomes
$$
n<8log n
$$
You can check that the function $f(x)=8log x-x$ has a maximum at $8log eapprox 3.47$, and so it is decreasing for $x>8log e$. Thus when you have found $nge 4$ satisfying the inequality, with $n+1$ not satisfying it, you're done.



We have $8log 6approx6.22>6$ and $8log 7approx6.76<7$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Forgive me for my question, but why did you use log$e$?
    $endgroup$
    – Evan Kim
    Dec 18 '18 at 0:45












  • $begingroup$
    @EvanKim Because the logarithm is in base 10.
    $endgroup$
    – egreg
    Dec 18 '18 at 7:44
















1












1








1





$begingroup$

You have a wrong simplification: the inequality $8n^2<64nlog n$ becomes
$$
n<8log n
$$
You can check that the function $f(x)=8log x-x$ has a maximum at $8log eapprox 3.47$, and so it is decreasing for $x>8log e$. Thus when you have found $nge 4$ satisfying the inequality, with $n+1$ not satisfying it, you're done.



We have $8log 6approx6.22>6$ and $8log 7approx6.76<7$.






share|cite|improve this answer









$endgroup$



You have a wrong simplification: the inequality $8n^2<64nlog n$ becomes
$$
n<8log n
$$
You can check that the function $f(x)=8log x-x$ has a maximum at $8log eapprox 3.47$, and so it is decreasing for $x>8log e$. Thus when you have found $nge 4$ satisfying the inequality, with $n+1$ not satisfying it, you're done.



We have $8log 6approx6.22>6$ and $8log 7approx6.76<7$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 17 '18 at 22:24









egregegreg

182k1486204




182k1486204












  • $begingroup$
    Forgive me for my question, but why did you use log$e$?
    $endgroup$
    – Evan Kim
    Dec 18 '18 at 0:45












  • $begingroup$
    @EvanKim Because the logarithm is in base 10.
    $endgroup$
    – egreg
    Dec 18 '18 at 7:44




















  • $begingroup$
    Forgive me for my question, but why did you use log$e$?
    $endgroup$
    – Evan Kim
    Dec 18 '18 at 0:45












  • $begingroup$
    @EvanKim Because the logarithm is in base 10.
    $endgroup$
    – egreg
    Dec 18 '18 at 7:44


















$begingroup$
Forgive me for my question, but why did you use log$e$?
$endgroup$
– Evan Kim
Dec 18 '18 at 0:45






$begingroup$
Forgive me for my question, but why did you use log$e$?
$endgroup$
– Evan Kim
Dec 18 '18 at 0:45














$begingroup$
@EvanKim Because the logarithm is in base 10.
$endgroup$
– egreg
Dec 18 '18 at 7:44






$begingroup$
@EvanKim Because the logarithm is in base 10.
$endgroup$
– egreg
Dec 18 '18 at 7:44




















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