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3 years ago

Which equation can be used to find the solution of (13^)d−5 =81 ?

Mathematics

2 answers:

DerKrebs [107]3 years ago

7 0

Use two main properties of powers:

1. $\left(\dfrac{1}{a}\right)^n=a^{-n};$

2. $a^n=a^m\Rightarrow n=m.$

Consider the equation

$\left(\dfrac{1}{3}\right)^{d-5}=81.$

Note that:

1. $\left(\dfrac{1}{3}\right)^{d-5}=3^{-1\cdot (d-5)}=3^{-d+5};$

2. $81=3^4.$

Then, the equation will take look

$3^{-d+5}=3^4.$

The bases of the left and right sides are the same, then the powers are the same too.

$-d+5=4.$

Answer: correct choice is C

Vadim26 [7]3 years ago

4 0

Your answer should be C.

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bija089 [108]

Part A:

The average rate of change refers to a function's slope. Thus, we are going to need to use the slope formula, which is:

$m = \dfrac{y_2 - y_1}{x_2 - x_1}$

$(x_1, y_1)$ and $(x_2, y_2)$ are points on the function

You can see that we are given the x-values for our interval, but we are not given the y-values, which means that we will need to find them ourselves. Remember that the y-values of functions refers to the outputs of the function, so to find the y-values simply use your given x-value in the function and observe the result:

$h(0) = 3(5)^0 = 3 \cdot 1 = 3$

$h(1) = 3(5)^1 = 3 \cdot 5 = 15$

$h(2) = 3(5)^2 = 3 \cdot 25 = 75$

$h(3) = 3(5)^3 = 3 \cdot 125 = 375$

Now, let's find the slopes for each of the sections of the function:

Section A

$m = \dfrac{15 - 3}{1 - 0} = \boxed{12}$

Section B

$m = \dfrac{375 - 75}{3 - 2} = \boxed{300}$

Part B:

In this case, we can find how many times greater the rate of change in Section B is by dividing the slopes together.

$\dfrac{m_B}{m_A} = \dfrac{300}{12} = 25$

It is 25 times greater. This is because $3(5)^x$ is an exponential growth function, which grows faster and faster as the x-values get higher and higher. This is unlike a linear function which grows or declines at a constant rate.