travis
  • travis
(logx)^(logx)=? pfft didn't do log for quite a while :P
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Mertsj
  • Mertsj
\[\log a ^{n}=nloga\]
travis
  • travis
not logx^(logx)
myininaya
  • 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

travis
  • travis
oh no, i have to proof its = x
myininaya
  • myininaya
what is the base?
myininaya
  • myininaya
i assumed it was 10
travis
  • travis
10 i think question didnt write
travis
  • travis
if it was another base im'm sure it wouldn't affect the answer
myininaya
  • myininaya
The base would just be some other number if not 10
myininaya
  • myininaya
\[(\log(x))^{\log(x)}=a^{\log(x) \cdot \log(\log(x))} \] if the base was a
travis
  • travis
Solve the equation (logx)^logx=x
travis
  • travis
thats how the question says
myininaya
  • myininaya
Oh we are solving an equation
travis
  • travis
log(logx)^logx=logx (logx)log(logx)=logx I cant cancel log on both sides right?
bahrom7893
  • bahrom7893
yea u can
myininaya
  • myininaya
\[\log(x)(\log(\log(x))-1)=0\]
myininaya
  • myininaya
subtract log(x) on both sides
myininaya
  • myininaya
both terms have log(x) in common
myininaya
  • myininaya
you factor
travis
  • travis
log(logx)=1 log(logx)=log10 logx=10 x=100 right?
myininaya
  • myininaya
what about x=10 for log(x)=0 since log(10)=1
travis
  • travis
oh wait log 100 isnt 10
travis
  • travis
x= 10^10!
myininaya
  • myininaya
\[x=10^{10}\]
myininaya
  • myininaya
you have two solutions
travis
  • travis
whats the other solution?
travis
  • travis
i took out logx so logx=0 right
travis
  • travis
or logx=1
myininaya
  • 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
  • myininaya
But if we check both solutions....
travis
  • 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
  • travis
cancel the 10 right?
Zarkon
  • Zarkon
I think you ment to say x=1 though after checking one sees that is not a solution
bahrom7893
  • bahrom7893
Travis you could have divided, you just made a mistake here: log(logx)=1 log(logx)=log10 logx=10 then x !=100
bahrom7893
  • bahrom7893
10^log(x) = 10^10 x = 10^10
myininaya
  • myininaya
oh yeah zarkon for some rason i solve log(x)=1
myininaya
  • myininaya
reason*
travis
  • travis
Yea I was too busy manipulating the log I didn't convert the log properly
myininaya
  • myininaya
log(x)=0 x=1 for sure! :)
travis
  • travis
yea so x =1 and x= 10^10
Zarkon
  • Zarkon
x=1 is only a solution if you define \[0^0=1\]
myininaya
  • myininaya
right which i don't define it that way
travis
  • travis
but if you sub x = 10^10 it works too
myininaya
  • myininaya
i would say that is the only solution
myininaya
  • myininaya
because after checking x=1 we get 0^0 like zarkon said
travis
  • travis
So reject that solution right

Looking for something else?

Not the answer you are looking for? Search for more explanations.