(logx)^(logx)=? pfft didn't do log for quite a while :P

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(logx)^(logx)=? pfft didn't do log for quite a while :P

Mathematics
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\[\log a ^{n}=nloga\]
not logx^(logx)
\[\text{ Let } y=(\log(x))^{\log(x)} \] Now do log( ) of both sides. \[\log(y)=\log[(\log(x))^{\log(x)}]\] Using properties of log we can rewrite as \[\log(y)=\log(x)\cdot \log(\log(x))\] Again we can use another property to rewrite this as Now I will write this as an exponential equation where the base is 10 assuming the base is 10 here. \[10^{\log(y)}=10^{\log(x)\cdot \log(\log(x))}\] \[y=10^{\log(x) \cdot \log(\log(x))}\] => \[(\log(x))^{\log(x)}=10^{\log(x) \cdot \log(\log(x))}\] Both still pretty ugly, but I honestly don't know what you want to show here.

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oh no, i have to proof its = x
what is the base?
i assumed it was 10
10 i think question didnt write
if it was another base im'm sure it wouldn't affect the answer
The base would just be some other number if not 10
\[(\log(x))^{\log(x)}=a^{\log(x) \cdot \log(\log(x))} \] if the base was a
Solve the equation (logx)^logx=x
thats how the question says
Oh we are solving an equation
log(logx)^logx=logx (logx)log(logx)=logx I cant cancel log on both sides right?
yea u can
\[\log(x)(\log(\log(x))-1)=0\]
subtract log(x) on both sides
both terms have log(x) in common
you factor
log(logx)=1 log(logx)=log10 logx=10 x=100 right?
what about x=10 for log(x)=0 since log(10)=1
oh wait log 100 isnt 10
x= 10^10!
\[x=10^{10}\]
you have two solutions
whats the other solution?
i took out logx so logx=0 right
or logx=1
\[(\log(x))^{\log(x)}=x\] D log( ) on both sides \[\log(x) \cdot \log(\log(x))=\log(x)\] Now subtract log(x) on both sides \[\log(x) \cdot \log(\log(x))-\log(x)=0\] Now factor the expression on the left side \[\log(x)(\log(\log(x))-1)=0\] Now set both factors =0 So we have \[\log(x)=0 \text{ or } \log(\log(x))-1=0\] So we should solve both equations \[x=10 \text{ or } \log(\log(x))=1\] \[x=10 \text{ or } 10^{\log(\log(x))}=10^1\] \[x=10 \text{ or } \log(x)=10\] \[x=10 \text{ or } 10^{\log(x)}=10^{10}\] \[x=10 \text{ or } x=10^{10}\]
But if we check both solutions....
I divided the log away instead of subtracting it,now i get where i was wrong, thanks a lot that was very kind of you!
cancel the 10 right?
I think you ment to say x=1 though after checking one sees that is not a solution
Travis you could have divided, you just made a mistake here: log(logx)=1 log(logx)=log10 logx=10 then x !=100
10^log(x) = 10^10 x = 10^10
oh yeah zarkon for some rason i solve log(x)=1
reason*
Yea I was too busy manipulating the log I didn't convert the log properly
log(x)=0 x=1 for sure! :)
yea so x =1 and x= 10^10
x=1 is only a solution if you define \[0^0=1\]
right which i don't define it that way
but if you sub x = 10^10 it works too
i would say that is the only solution
because after checking x=1 we get 0^0 like zarkon said
So reject that solution right

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