How do you solve ${log}_{e} x + {({log}_{e} x)}^{2} = 6$ $log_e x + (log_e x)^2 = 6$ ?

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Answer 1 · 2018-05-28T13:23:40

$x = e^{2}$ $x=e^2$

With $t = ln (x)$ $t=ln(x)$ we get

$t^{2} + t - 6 = 0$ $t^2+t-6=0$

solving this equation

$t_{1} = 2$ $t_1=2$

or

$t_{2} = - 3$ $t_2=-3$ (which is not a solution!)

so

$x = e^{2}$ $x=e^2$

How do you solve log_e x + (log_e x)^2 = 6logex+(logex)2=6log_e x + (log_e x)^2 = 6?