Solving exponential equation in one variable.

Question

I'm stuck with solving this exponential equation: $(1-v)^v = 2$

PS. I'm sorry I can show no solution that I could try myself.

Answer 1

Convert to $$ v \ln (1-v) = \ln (2). $$ For $0 < v < 1$, the log will be negative, but $v > 0$, so there's no solution in that interval. For $1 - v < 0$ (i.e., $v > 1$) it's not clear what any power of $v$ except for even integer powers actually means. So the only hope is $1 - v > 0$, i.e., $v < 1$, and then by the first case, this becomes $v < 0$.

In that case, $\ln(1-v)$ will be positive (it's larger than $\ln 1 = 0$, because $v$ is negative, but $v$ is negative, so the product is negative.

In short: this appears to have no solutions.

Answer 2

$\exp(v \log(1-v)) = 2$, $v \log(1-v) = \log(2)$

you have to study the function $f:x\rightarrow x\log(1-x)$ for $x<1$.

Answer 3

Since you marked it real analysis: if $v>0$, then $(1-v)<1$ and so $(1-v)^x<1$ for any $x>0$ (in particular, for $x=v$). If $v\leq0$, then $1-v\geq1$, and $(1-v)^{-x}\leq1$ for any $x\geq0$ (in particular, for $-x=v$). Hence, no solutions exist.