Showing there exists a solution to $x \ln{x} = 1$ in $x \in [1.5, 2]$

Question

This was given as a bonus question on my last quiz, and I couldn't figure it out. We just learned about Newton's method and finite differentiation methods but I don't see how any of that applies to this nonlinear equation... any ideas?

Answer 1

Hint: $1.5^{1.5} < 1.5^2 = 2.25 < e < 2^2 = 4$

Answer 2

A variant:

$\ln 1.5=\ln\Bigl(1+\dfrac12\Bigr)<\dfrac12$, hence $\;1.5\ln1.5<\dfrac34$,

while $\;2>\sqrt{\mathrm e}$, hence $\;2\ln 2>2\cdot\dfrac12=1$.

Answer 3

third variant.

$e < 4$

$e^{1/2} < 2$

$\frac 1 2 < \ln 2$

$x \ln x |_{x=2} = 2 \ln 2 > 2*\frac 1 2 = 1$

$e > 2$

$e^2 > 4 > \frac {27}{8}$

$e^{\frac 2 3} > (\frac {27}{8})^{\frac 1 3} = 1.5$

$ \frac 2 3 > \ln 1.5$

$x ln x|_{x = 1.5} = 1.5 \ln 1.5 < \frac 3 2 \frac 2 3 = 1$

So $x \ln x = 1$ for some value in $[1.5, 2]$ via intermediate value theorem.