What is the f(z)=log_{3}(z)+log_{9}(z-2) ?

The f(z)=log_{3}(z)+log_{9}(z-2) is

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f(z)=log_{3}(z)+log_{9}(z-2)

f (z) = lo g_{3} (z) + lo g_{9} (z - 2)

Domain : z > 2

X Intercepts : (2.20556 \dots, 0)

Asymptotes : Vertical z = 2

Extreme Points : None

Interval Notation

Domain : (2, \infty)

Solution steps

Domain of

lo g_{3} (z) + lo g_{9} (z - 2) : z > 2

Axis interception points of

lo g_{3} (z) + lo g_{9} (z - 2) :

X Intercepts

: (2.20556 \dots, 0)

Asymptotes of

lo g_{3} (z) + lo g_{9} (z - 2) :

Vertical

: z = 2

Extreme Points of

lo g_{3} (z) + lo g_{9} (z - 2) :

None

y = - (t - 1)^{2} + 38

g (x) = - 5^{x} + 5

y = - 30000 x + 150000

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

f (x) = ln (3.5 x + 1.1) - 1.2