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

- travis

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

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- Mertsj

\[\log a ^{n}=nloga\]

- travis

not logx^(logx)

- myininaya

\[\text{ Let } y=(\log(x))^{\log(x)} \]
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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## More answers

- travis

oh no, i have to proof its = x

- myininaya

what is the base?

- myininaya

i assumed it was 10

- travis

10 i think
question didnt write

- travis

if it was another base im'm sure it wouldn't affect the answer

- myininaya

The base would just be some other number if not 10

- myininaya

\[(\log(x))^{\log(x)}=a^{\log(x) \cdot \log(\log(x))} \]
if the base was a

- travis

Solve the equation (logx)^logx=x

- travis

thats how the question says

- myininaya

Oh we are solving an equation

- travis

log(logx)^logx=logx
(logx)log(logx)=logx
I cant cancel log on both sides right?

- bahrom7893

yea u can

- myininaya

\[\log(x)(\log(\log(x))-1)=0\]

- myininaya

subtract log(x) on both sides

- myininaya

both terms have log(x) in common

- myininaya

you factor

- travis

log(logx)=1
log(logx)=log10
logx=10
x=100 right?

- myininaya

what about x=10 for log(x)=0
since log(10)=1

- travis

oh wait log 100 isnt 10

- travis

x= 10^10!

- myininaya

\[x=10^{10}\]

- myininaya

you have two solutions

- travis

whats the other solution?

- travis

i took out logx so logx=0 right

- travis

or logx=1

- myininaya

\[(\log(x))^{\log(x)}=x\]
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
\[x=10 \text{ or } \log(\log(x))=1\]
\[x=10 \text{ or } 10^{\log(\log(x))}=10^1\]
\[x=10 \text{ or } \log(x)=10\]
\[x=10 \text{ or } 10^{\log(x)}=10^{10}\]
\[x=10 \text{ or } x=10^{10}\]

- myininaya

But if we check both solutions....

- travis

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!

- travis

cancel the 10 right?

- Zarkon

I think you ment to say x=1
though after checking one sees that is not a solution

- bahrom7893

Travis you could have divided, you just made a mistake here:
log(logx)=1
log(logx)=log10
logx=10
then x !=100

- bahrom7893

10^log(x) = 10^10
x = 10^10

- myininaya

oh yeah zarkon for some rason i solve log(x)=1

- myininaya

reason*

- travis

Yea I was too busy manipulating the log I didn't convert the log properly

- myininaya

log(x)=0
x=1 for sure! :)

- travis

yea so x =1 and x= 10^10

- Zarkon

x=1 is only a solution if you define \[0^0=1\]

- myininaya

right which i don't define it that way

- travis

but if you sub x = 10^10 it works too

- myininaya

i would say that is the only solution

- myininaya

because after checking x=1 we get 0^0 like zarkon said

- travis

So reject that solution right

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