What is the y=log_{10}(((x-5))/((x+2))) ?

The y=log_{10}(((x-5))/((x+2))) is

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y=log_{10}(((x-5))/((x+2)))

y = lo g_{10} (\frac{( x - 5 )}{( x + 2 )})

Domain : x < - 2 or x > 5

Range : f (x) < 0 or f (x) > 0

Intercepts : None

Asymptotes : Vertical x = - 2, x = 5

Extreme Points : None

Interval Notation

Domain : (- \infty, - 2) \cup (5, \infty)

Range : (- \infty, 0) \cup (0, \infty)

Solution steps

Domain of

lo g_{10} (\frac{x - 5}{x + 2}) : x < - 2 or x > 5

Range of

lo g_{10} (\frac{x - 5}{x + 2}) : f (x) < 0 or f (x) > 0

Axis interception points of

lo g_{10} (\frac{x - 5}{x + 2}) :

None

Asymptotes of

lo g_{10} (\frac{x - 5}{x + 2}) :

Vertical

: x = - 2, x = 5

Extreme Points of

lo g_{10} (\frac{x - 5}{x + 2}) :

None

f (x) = \frac{1}{3 \cdot x + 3}

f (x) = \frac{( x ^{2} + 1 )}{( 2 x )}

f (x) = 3 lo g_{4} (x) + 5 lo g_{2} (x) - 14

f (x) = - x (- 1 x - 2) (3 x + 3) 2 (- 1 x - 6) 3

f (x) = 15 x^{3} + 3 ln (x)