How to Solve the Equation (x)^x^3=3?

[John Creighto]

No. I meant to write

$$\left (\sqrt 2 ^{\sqrt 2}\right)^{\sqrt 2} = 2$$

$$\left( \left(\sqrt[3]3 ^{\sqrt[3]3}\right)^{\sqrt[3]3}\right)^{\sqrt[3]3} = 3$$

I left out the parentheses, which are absolutely essential as exponentiation is non-associative.

No, you put them in but apparently in the wrong spot.

 

[sutupidmath]

$$\log_n\Biggl(\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_{n+1}\Biggr) =$$

probbably he meant to write it sth like this:

$$\log_n (...(((\sqrt[n]^n)^{\sqrt [n]^n})^{\sqrt [n]^n})^{\sqrt [n]^n}...)^{\sqrt [n]^n} =\sqrt[n]^n \log_n (...(((\sqrt[n]^n)^{\sqrt [n]^n})^{\sqrt [n]^n})^{\sqrt [n]^n}...)$$=$$\sqrt [n]^n^{2} \log_n(..((\sqrt[n]^n)^{\sqrt[n]^n})^{\sqrt[n]^n})..)$$..=...=$$\sqrt[n]^n^{n}log_n\sqrt[n]^n=n*\frac{1}{n}log_n n=1$$ and hence:



$$(...(((\sqrt[n]^n)^{\sqrt [n]^n})^{\sqrt [n]^n})^{\sqrt [n]^n}...)^{\sqrt [n]^n}=n$$

Does this make any sense??

 

[John Creighto]

$$\log_n\Biggl(\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_{n+1}\Biggr) =$$

probbably he meant to write it sth like this:

$$\log_n (...(((\sqrt[n]^n)^{\sqrt [n]^n})^{\sqrt [n]^n})^{\sqrt [n]^n}...)^{\sqrt [n]^n} =\sqrt[n]^n \log_n (...(((\sqrt[n]^n)^{\sqrt [n]^n})^{\sqrt [n]^n})^{\sqrt [n]^n}...)$$=$$\sqrt [n]^n^{2} \log_n(..((\sqrt[n]^n)^{\sqrt[n]^n})^{\sqrt[n]^n})..)$$..=...=$$\sqrt[n]^n^{n}log_n\sqrt[n]^n=n*\frac{1}{n}log_n n=1$$ and hence:



$$(...(((\sqrt[n]^n)^{\sqrt [n]^n})^{\sqrt [n]^n})^{\sqrt [n]^n}...)^{\sqrt [n]^n}=n$$

Does this make any sense??

You don't even need logs to show that sense

(x^a)^b=x^(ab)