With every multiple utterance of "log," I kept anticipating you'd follow with "it's big, it's heavy, it's wood." However, I realize this is an obscure cultural reference.
if you're on the website, after entering input, just click the settings icon on top, then click on the color, and you'll see three option: normal line, dotted line, dotted line but more dotty. choose ur option
Thank you! Graph your function in desmos, click and hold your mouse on the circle until the 'Lines' menu comes up (see image) and then pick the dotted line: postimg.cc/qN6DVznQ
Not so difficult. Put b=log(x), and a=Expression Then a=10^b^(log(log(b))/log(b)) log(a)=b^(log(log(b))/log(b)) log(log(a))=(log(log(b))/log(b))*log(b) log(log(a))=log(log(b)) a=b a=log(x) Expression=log(x) Syber, am my solution right? Plese help me check it. Thank you. 😊😊😊😊😊😊
10^(log x)^((log (log (log x)))/(log (log x))) I set this expression = k. Then I set c = log (log x)). Then log both sides. log k = log (10^(log x)^(log c)/c log k = (log x)^(log c)/c log (log k) = log (log x)^(log c / c) log (log k) = (log c / c) * log (log x) log (log k) = (log c / c) * c log (log k) = log c log k = c log k = log (log x) k = log x
In the first method he obtains log(t) because of the simplification, but t is equal to log(x), so in this first method we get log(log(x)) instead of log(x). Is that correct?
10^(log(x))^[(log(log(log(x))/log(log(x))] Well at first we can let: Log(log(log(x)))=u Then we can see ; Log(log(x))=10^(u) And log(x)=10^(10^(u)) And just placement! So we have ; 10^(10^(10^u))^[u/10^u] And simplify!( only ) 10^(10^u) = log(x) After answering, we should celebrate, right?!🥳(If we have solved it correctly) The End
Third method using the base change formula: log _base_a (b) = log _base_c (b)/log_base_c (a)
10^(logx)^(((log(log(logx))/log(logx)) = 10^(logx)^(log_base_logx (log(logx)) = 10^(log(logx)) = logx
Nice
"lets start with the 2nd one"
Yeah that made me laugh
🤪
These are in "professor order"
Great approach! I solved it by naming all the expression “y” and logging both sides twice!! Finally you get y=logx❤
Excellent!
@@SyberMath thank you!!!!!!❤️❤️❤️
Wonderful! Thank you.
Glad you enjoyed it!
With every multiple utterance of "log," I kept anticipating you'd follow with "it's big, it's heavy, it's wood."
However, I realize this is an obscure cultural reference.
😜
Nice vid , but how did you do the dotted line in desmos?
if you're on the website, after entering input, just click the settings icon on top, then click on the color, and you'll see three option: normal line, dotted line, dotted line but more dotty. choose ur option
Thank you!
Graph your function in desmos, click and hold your mouse on the circle until the 'Lines' menu comes up (see image) and then pick the dotted line:
postimg.cc/qN6DVznQ
Nice! I learned something new 🤩
Log x; x>10
Good point
log x, easy
Very cool.
Change of variable
First method -> second method
Second method -> first method.
Not so difficult.
Put b=log(x), and a=Expression
Then
a=10^b^(log(log(b))/log(b))
log(a)=b^(log(log(b))/log(b))
log(log(a))=(log(log(b))/log(b))*log(b)
log(log(a))=log(log(b))
a=b
a=log(x)
Expression=log(x)
Syber, am my solution right? Plese help me check it. Thank you. 😊😊😊😊😊😊
10^(log x)^((log (log (log x)))/(log (log x)))
I set this expression = k. Then I set c = log (log x)). Then log both sides.
log k = log (10^(log x)^(log c)/c
log k = (log x)^(log c)/c
log (log k) = log (log x)^(log c / c)
log (log k) = (log c / c) * log (log x)
log (log k) = (log c / c) * c
log (log k) = log c
log k = c
log k = log (log x)
k = log x
In the first method he obtains log(t) because of the simplification, but t is equal to log(x), so in this first method we get log(log(x)) instead of log(x). Is that correct?
10^(log(x))^[(log(log(log(x))/log(log(x))]
Well at first we can let:
Log(log(log(x)))=u
Then we can see ;
Log(log(x))=10^(u)
And log(x)=10^(10^(u))
And just placement!
So we have ;
10^(10^(10^u))^[u/10^u]
And simplify!( only )
10^(10^u) = log(x)
After answering, we should celebrate, right?!🥳(If we have solved it correctly)
The End
Log(x)
logx