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

not logx^(logx)

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

But if we check both solutions....

cancel the 10 right?

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

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