What is 2log_{5}(x)-log_{x}(125)<1 ?

The solution to 2log_{5}(x)-log_{x}(125)<1 is 0<x< 1/5 \lor 1<x<5sqrt(5)

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2log_{5}(x)-log_{x}(125)<1

2 lo g_{5} (x) - lo g_{x} (125) < 1

0 < x < \frac{1}{5} or 1 < x < 55

Interval Notation

(0, \frac{1}{5}) \cup (1, 55)

Decimal Notation

0 < x < 0.2 or 1 < x < 11.18033 \dots

Solution steps

2 lo g_{5} (x) - lo g_{x} (125) < 1

Apply log rules

2 lo g_{5} (x) - lo g_{5} (12 5^{\frac{1}{l o g _{5} ( x )}}) < 1

Apply log rule

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x),

assuming

x \geq 0

2 lo g_{5} (x) - lo g_{5} (125) \frac{1}{lo g _{5} ( x )}

Simplify

lo g_{5} (125) : 3

2 lo g_{5} (x) - 3 \frac{1}{lo g _{5} ( x )}

Simplify

2 lo g_{5} (x) - 3 \cdot \frac{1}{lo g _{5} ( x )} : 2 lo g_{5} (x) - \frac{3}{lo g _{5} ( x )}

2 lo g_{5} (x) - \frac{3}{lo g _{5} ( x )} < 1

Let

u = lo g_{5} (x)

2 u - \frac{3}{u} < 1

2 u - \frac{3}{u} < 1 : u < - 1 or 0 < u < \frac{3}{2}

u < - 1 or 0 < u < \frac{3}{2}

Substitute back

u = lo g_{5} (x)

lo g_{5} (x) < - 1 or 0 < lo g_{5} (x) < \frac{3}{2}

lo g_{5} (x) < - 1 : 0 < x < \frac{1}{5}

0 < lo g_{5} (x) < \frac{3}{2} : 1 < x < 55

Combine the intervals

0 < x < \frac{1}{5} or 1 < x < 55

- 69 x^{2} + 4050 x = 250 x + 5080

6 x^{2} + 23 x + 21 = 0

6 x 1 + 3 x 2 + 5 x 3 + 2 x 4 \leq 10

- 3 x - 11 = - 2

- \frac{64}{4}

What is 2log_{5}(x)-log_{x}(125)<1 ?
The solution to 2log_{5}(x)-log_{x}(125)<1 is 0<x< 1/5 \lor 1<x<5sqrt(5)