I was wondering whether there is a formula for solving $x^x = N$ for $x$. I thought I could just take logs, but the problem is that there are still two $x$'s and I need to express everything for one $x$ in terms of $N$.
Thanks.
I was wondering whether there is a formula for solving $x^x = N$ for $x$. I thought I could just take logs, but the problem is that there are still two $x$'s and I need to express everything for one $x$ in terms of $N$.
Thanks.
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