Log inequalities

Question

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

Answer 1

The inequality you've got looks correct (presumably it comes from exponentiating both sides, then dividing by $5^{1+1/(2x)}$).

Hint to continue from here: write $y=5^{1-1/(2x)}$ and rearrange to get an equation in $y$.

Answer 2

By the logarithm rule we get $$\log(5^{1/x}+5^3)<\log(6\cdot 5^{1+1/2x})$$ so we get

$$5^{1/x}+5^3<6\cdot 5^{1+1/2x}$$

Can you finish?

Answer 3

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

$$5<5^{\frac{1}{2x}}<25=5^2$$

$$1<\frac{1}{2x}<2 \iff \frac12 <2x <1 \iff \frac 14 < x < \frac 12$$