Suppose you are given the relation that [math] a^a= b [/math] . Then how you do you proceed to find a in terms of b ? Or, the solution doesn't exist.

Suppose you are given the relation that [math] a^a= b [/math] . Then how you do you proceed to find a in terms of b ? Or, the solution doesn't exist.

Transcendental equations

Irrelevant link.

 

[math]a^a=b \Rightarrow a=\log_a b =(^a\sqrt{b})[/math]. Of course this generalizes - [math]x^y=z \Rightarrow y=log_xz\,\,\ne \,\,(^y\sqrt{z})[/math]

Irrelevant link.

 

[math]a^a=b \Rightarrow a=\log_a b =(^a\sqrt{b})[/math].

The link is fully relevant, he's trying to find "a" as a function of "b". You can't do that.

 

 

 

Of course this generalizes - [math]x^y=z \Rightarrow y=log_xz\,\,\ne \,\,(^y\sqrt{z})[/math]

 

This is a DIFFERENT problem. Do you see the difference?

Edited March 28, 20178 yr by zztop