Question

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

Answer 1

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

Answer 2

not logx^(logx)

Answer 3

\[\text{ Let } y=(\log(x))^{\log(x)} \]
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.

Answer 4

oh no, i have to proof its = x

Answer 5

what is the base?

Answer 6

i assumed it was 10

Answer 7

10 i think
question didnt write

Answer 8

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

Answer 9

The base would just be some other number if not 10

Answer 10

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

Answer 11

Solve the equation (logx)^logx=x

Answer 12

thats how the question says

Answer 13

Oh we are solving an equation

Answer 14

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

Answer 15

yea u can

Answer 16

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

Answer 17

subtract log(x) on both sides

Answer 18

both terms have log(x) in common

Answer 19

you factor

Answer 20

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

Answer 21

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

Answer 22

oh wait log 100 isnt 10

Answer 23

x= 10^10!

Answer 24

\[x=10^{10}\]

Answer 25

you have two solutions

Answer 26

whats the other solution?

Answer 27

i took out logx so logx=0 right

Answer 28

or logx=1

Answer 29

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

Answer 30

But if we check both solutions....

Answer 31

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!

Answer 32

cancel the 10 right?

Answer 33

I think you ment to say x=1

though after checking one sees that is not a solution

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

reason*

Answer 38

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

Answer 39

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

Answer 40

yea so x =1 and x= 10^10

Answer 41

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

Answer 42

right which i don't define it that way

Answer 43

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

Answer 44

i would say that is the only solution

Answer 45

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

Answer 46

So reject that solution right