Solve each of the following equations for x. a) ln(5x)=0 x=? b) log_5 (2x)+log_5 (4x)=2 x=?

Solve each of the following equations for x.
a) $\mathrm{ln}\left(5x\right)=0\phantom{\rule{0ex}{0ex}}x=?$
b) ${\mathrm{log}}_{5}\left(2x\right)+{\mathrm{log}}_{5}\left(4x\right)=2\phantom{\rule{0ex}{0ex}}x=?$
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Nigerkamg5
Suppose $q={x}^{2}+2x$ is rational.
Then, by the quadratic formula, $x=\frac{-2±\sqrt{4+4q}}{2}=-1±\sqrt{\alpha }$, where $\alpha =q+1$ can be any rational number that is not a perfect square (since x is irrational).
Now,
${x}^{3}-6x=\left(-1±\sqrt{\alpha }{\right)}^{3}-6\left(-1±\sqrt{\alpha }\right)=±{\alpha }^{3/2}-3\alpha \mp 3\sqrt{\alpha }+5$
will be rational if and only if ${\alpha }^{3/2}-3\sqrt{\alpha }$ is (since the other terms are always rational). But this is equal to $\sqrt{\alpha }\left(\alpha -3\right)$. Since $\alpha -3$ is rational and $\sqrt{\alpha }$ irrational, the entire quantity will be rational iff it is zero: that is, iff $\alpha =3$.
Thus the numbers $x$ that satisfy the condition are $-1±\sqrt{3}$
Uroskopieulm