What is log_{0.5}(7x+5)<= log_{0.5}(5x-3) ?

The solution to log_{0.5}(7x+5) 3/5

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log_{0.5}(7x+5)<= log_{0.5}(5x-3)

lo g_{0.5} (7 x + 5) \leq lo g_{0.5} (5 x - 3)

x > \frac{3}{5}

Interval Notation

(\frac{3}{5}, \infty)

Decimal Notation

x > 0.6

Solution steps

lo g_{0.5} (7 x + 5) \leq lo g_{0.5} (5 x - 3)

Apply log rules

7 x + 5 \geq 5 x - 3 and 5 x - 3 > 0

7 x + 5 \geq 5 x - 3 and 5 x - 3 > 0 : x > \frac{3}{5}

x > \frac{3}{5}

3 ∣ 2 x - 9 ∣ - 5 < 16

2 = - 35 \cdot (2 \cdot \frac{x}{- 0.00796} - (\frac{x}{- 0.00796})^{2})

12 x^{2} - 7 x - 10

(- 125)^{- \frac{1}{3}}

- 5 + 9

What is log_{0.5}(7x+5)<= log_{0.5}(5x-3) ?
The solution to log_{0.5}(7x+5)<= log_{0.5}(5x-3) is x> 3/5