May anyone explain to me how do I solve this, /sci/?
>>7734007
Try expressing it in a Banach space first, then it should be trivial.
>>7734011
This
x^log_x(a) = a
raise both sides over x
x^logx(9) = x^(2/3)
left side cancels
9 = x^(2/3)
raise both sides to power 3/2
9^(3/2) = x^(2/3)(3/2) = x^1 = x
so
x=27
>>7734007
Sure thing.
Note that the problem is asking for this:
x, the base, raised to the 2/3 is equal to 9.
x^(2/3)=9
x^((2/3)*3)=9^3
x^2=729
x=27
Thus, The base is 27.
>>7734007
Rewrite it as 9=x^(2/3), square root both sides, then cube both sides. You get x=27.
>>7734036
It's an easy problem why not just help him and let him go back to doing other stuff?
>>7734040
>it's an easy problem
Exactly, assuming OP doesn't have a fucking degenerative disorder all he has to do is flip open his notes, Google, or where ever the fuck he learned the material from and get the answer in less time than it would of taken to make the thread and fish responses.
>>7734026
>>7734028
>>7734031
>>7734032
Thanks, guys. You're the best.
>>7734052
OP here. Tried looking for some, including Google and the book I'm studying from but neither had it as well-explained as >>7734031
Did. I haven't touched math about 3-4 years because of mandatory army service, so I still need some help here and there.
27
x=27
>>7734007
1/log_9(x)=2/3
log_9(x)=3/2
x=9^3/2
x=27
>>7734007
[math] \displaystyle log_ba = \frac{log_ca}{log_cb} [/math]
[math] log_x(9) = \frac{ln(9)}{ln(x)} [/math]
Therefore, you can easily obtain x = 27.
log_a(x)=y<=>a^y=x
therefore
x^2/3=9
x^2=9^3
x^2=3^6
x=3^6/2
x=3^3
x=27
do your own homework, reported nigga
>>7735292
>not using log base 9
log base x of 9 = 2/3
x^(2/3) = 9
x^(2/3)^(3/2) = 9^(3/2)
x = 9^(3/2)
x = 3^3
x = 27
27.
