Question

$\large \bold \red{(x + 1) {}^{x} {e}^{x + 1} = ( \frac{ - x}{e} ) {}^{x} }$​

Answer 1

$\large \bold \red{(x + 1) {}^{x} {e}^{x + 1} = ( \frac{ - x}{e} ) {}^{x} }$

${\large \bold \red{ \bigg((x + 1) {}^{x} {e}^{x + 1} \bigg) {}^{ \frac{1}{x} } = \bigg ( (\frac{ - x}{e} ) {}^{x}\bigg) {}^{ \frac{1}{x} } }}$

$\large\bold \red{ \frac{(x + 1)}{x} e^{1 + \frac{1}{x} } = \cancel \frac{1}{x} - \frac{ \cancel{x}}{e} }$

$\bold \red{W \bigg((1 + \frac{1}{x} )e {}^{1 + \frac{1}{x} } \bigg) = W - \frac{1}{e} \bigg)}$

$\large \bold \red{1 + \frac{1}{x} = - 1}$

$\large \bold \red{x = \frac{ - 1}{2} }$

Answer 2

Answer:

$\large \bold \red{(x + 1) {}^{x} {e}^{x + 1} = ( \frac{ - x}{e} ) {}^{x} }$

${\large \bold \red{ \bigg((x + 1) {}^{x} {e}^{x + 1} \bigg) {}^{ \frac{1}{x} } = \bigg ( (\frac{ - x}{e} ) {}^{x}\bigg) {}^{ \frac{1}{x} } }}$

$\large\bold \red{ \frac{(x + 1)}{x} e^{1 + \frac{1}{x} } = \cancel \frac{1}{x} - \frac{ \cancel{x}}{e} }$

$\bold \purple{W \bigg((1 + \frac{1}{x} )e {}^{1 + \frac{1}{x} } \bigg) = W - \frac{1}{e} \bigg)}$

$\large \bold \purple{1 + \frac{1}{x} = - 1}$

$\large \bold \purple{x = \frac{ - 1}{2} }$