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dusya [7]
3 years ago
14

What value of x satisfies this equation? log(5x+20)=2

Mathematics
2 answers:
Serggg [28]3 years ago
7 0

Answer:

16 was right

Step-by-step explanation:

Ipatiy [6.2K]3 years ago
5 0

Answer:

16

Step-by-step explanation:

Given expression:

       log(5x + 20) = 2

From the given expression, we are to find x;

   Using the rule of logarithm, we can solve this problem;

             logₐb = C

                    b  = Cᵃ

So;

            log(5x + 20) = 2

  this is to base 10;

                    5x + 20 = 10²

                    5x  = 100 - 20

                    5x  = 80

                      x  = 16

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<img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cfrac%7B%282x%2B1%29%28x-5%29%7D%7B%28x-5%29%28x%2B4%29%5E%7B2%7D%20%7D" id
dybincka [34]

i) The given function is

f(x)=\frac{(2x+1)(x-5)}{(x-5)(x+4)^2}

The domain is

(x-5)(x+4)^2\ne 0

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ii) For vertical asymptotes, we simplify the function to get;

f(x)=\frac{(2x+1)}{(x+4)^2}

The vertical asymptote occurs at

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2x+1=0

2x=-1

x=-\frac{1}{2}

iv) To find the y-intercept, we substitute x=0 into the reduced fraction.

f(0)=\frac{(2(0)+1)}{(0+4)^2}

f(0)=\frac{(1)}{(4)^2}

f(0)=\frac{1}{16}

v) The horizontal asymptote is given by;

lim_{x\to \infty}\frac{(2x+1)}{(x+4)^2}=0

The horizontal asymptote is y=0.

vi) The function has a hole at x-5=0.

Thus at x=5.

This is the factor common to both the numerator and the denominator.

vii) The function is a proper rational function.

Proper rational functions do not have oblique asymptotes.

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