Log inequalities Solve the inequality
$\log(5^{1/x} +5^3) < \log 6 + \log 5^{1+\frac{1}{2x}}$
I came up with an equation  
$5^{\frac{4x-1}{2x}} + 5^{\frac{-2x+1}{2x}}-6<0$
Which I couldn't solve, i tried to make it quadratic but I couldn't

Yes, since $\log$ function is monotonic increasing we have
$$\log(5^{1/x}+5^3)<\log(6\cdot 5^{1+1/2x})\iff 5^{1/x}+5^3<6\cdot 5^{1+1/2x}$$
that is
$$5^{1/x-1-1/2x}+5^{3-1-1/2x}<6$$
and then your result

$$5^{\frac{4x-1}{2x}}+5^{\frac{1-2x}{2x}}-6<0$$

then
$$5^{2-\frac{1}{2x}}+5^{\frac{1}{2x}-1}-6<0$$
$$25\cdot 5^{-\frac{1}{2x}}+\frac15 \cdot 5^{\frac{1}{2x}}-6<0$$
that is by $y=5^{\frac{1}{2x}}$
$$\frac{25}y+\frac y 5-6<0 \implies y^2-30y+125<0$$
$$y=\frac{30\pm\sqrt{900-500}}{2}=15\pm 10 \implies 5<y<25$$
that is, observing that exponential function is monotonic increasing
$$5<5^{\frac{1}{2x}}<25=5^2$$
$$1<\frac{1}{2x}<2 \iff \frac12 <2x <1 \iff \frac 14 < x < \frac 12$$
