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lilavasa [31]
3 years ago
11

What is the logarithmic function modeled by the following table?

Mathematics
2 answers:
Vikki [24]3 years ago
7 0
The answer would be the following: f(x)=5^x
sleet_krkn [62]3 years ago
6 0
So we need to come up with a function that when we plug x into it, it gives us    f(x). Logarithims ask: what exponent do I need to put on b (see the diagram to see what b is) to make it equal x. In the diagram below y is our f(x).

So we have logb (x) = f(x), we need to find what b is. Well the values look like 5 then 5 squared, then 5 cubed etc. So b=5.

To test this put one of the exponents (f(x)-values) on b and see if it equals the x value next to it in the table.

The answer is:

log 5 x = f(x)

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Consider the two functions:
koban [17]

Answer:

a) The x value of the point where the two equations intersect in terms of a is x=\frac{40}{4+5a}

b) The value of the functions at the point where they intersect is \frac{10 (28 + 15 a)}{4 + 5 a}

c) The partial derivative of f with respect to x is \frac{\partial f}{\partial x} = -5a and the partial derivative of f with respect to a is \frac{\partial f}{\partial x} = -5x

d) The value of \frac{\partial f}{\partial x}(3,2) = -10 and \frac{\partial f}{\partial a}(3,2) = -15

e) \upsilon_1=-\frac{3}{4} = -0.75 and \upsilon_2=-\frac{3}{4} = -0.75

f) equation \upsilon_1 = \frac{-5a\cdot x}{70-5ax}=\frac{ax}{ax-14} and \upsilon_2 = \frac{-5a\cdot a}{70-5ax}=\frac{a^2}{ax-14}

Step-by-step explanation:

a) In order to find the x we just need to equal the equations and solve for x:

f(x,a)=g(x)\\70-5xa = 30+4x\\70-30 = 4x+5xa\\40 = x(4+5a)\\\boxed {x = \frac{40}{4+5a}}

b) Since we need to find the value of the function in the intersection point we just need to substitute the result from a) in one of the functions. As a sanity check , I will do it in both and the value (in terms of a) must be the same.

f(x,a)=70-5ax\\f(\frac{40}{4+5a}, a) = 70-5\cdot a \cdot  \frac{40}{4+5a}\\f(\frac{40}{4+5a}, a) = 70 - \frac{200a}{4+5a}\\f(\frac{40}{4+5a}, a) = \frac{70(4+5a) -200a}{4+5a}\\f(\frac{40}{4+5a}, a) =\frac{280+350a-200a}{4+5a}\\\boxed{ f(\frac{40}{4+5a}, a) =\frac{10(28+15a)}{4+5a}}

and for g(x):

g(x)=30+4x\\g(\frac{40}{4+5a})=30+4\cdot \frac{40}{4+5a}\\g(\frac{40}{4+5a})=\frac{30(4+5a)+80}{4+5a}\\g(\frac{40}{4+5a})=\frac{120+150a+80}{4+5a}\\\boxed {g(\frac{40}{4+5a})=\frac{10(28+15a)}{4+5a}}

c) \frac{\partial f}{\partial x} = (70-5xa)^{'}=70^{'} - \frac{\partial (5xa)}{\partial x}=0-5a\\\frac{\partial f}{\partial x} =-5a

\frac{\partial f}{\partial a} = (70-5xa)^{'}=70^{'} - \frac{\partial (5xa)}{\partial a}=0-5x\\\frac{\partial f}{\partial a} =-5x

d) Then evaluating:

\frac{\partial f}{\partial x} =-5a\\\frac{\partial f}{\partial x} =-5\cdot 2=-10

\frac{\partial f}{\partial a} =-5x\\\frac{\partial f}{\partial a} =-5\cdot 3=-15

e) Substituting the corresponding values:

\upsilon_1 = \frac{\partial f(3,2)}{\partial x}\cdot \frac{3}{f(3,2)} \\\upsilon_1 = -10 \cdot \frac{3}{40}  = -\frac{3}{4} = -0.75

\upsilon_2 = \frac{\partial f(3,2)}{\partial a}\cdot \frac{3}{f(3,2)} \\\upsilon_2 = -15 \cdot \frac{2}{40}  = -\frac{3}{4} = -0.75

f) Writing the equations:

\upsilon_1=\frac{\partial f (x,a)}{\partial x}\cdot \frac{x}{f(x,a)}\\\upsilon_1=-5a\cdot \frac{x}{70-5xa}\\\upsilon_1=\frac{-5ax}{70-5ax}=\frac{-5ax}{-5(ax-14)}\\\boxed{\upsilon_1=\frac{ax}{ax-14} }

\upsilon_2=\frac{\partial f (x,a)}{\partial x}\cdot \frac{a}{f(x,a)}\\\upsilon_2=-5a\cdot \frac{a}{70-5xa}\\\upsilon_2=\frac{-5a^2}{70-5ax}=\frac{-5a^2}{-5(ax-14)}\\\boxed{\upsilon_2=\frac{a^2}{ax-14} }

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Sindrei [870]

Answer:

100%

Step-by-step explanation:

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PLEASE HELP,, MARKING BRAINLIEST!!!<br><br> Given coordinates N(-1,-11) and P(-1,-3), find NP?
Allushta [10]

Answer:

its 8 units long

Step-by-step explanation:

since x is the same we dont have to do much since its a straight line up and down

find the difference in the ys

-11 - -3 = -8

8 units long

hope this helps <3

4 0
3 years ago
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