Solve the exponential equation 9^2x = 27

Mathematics

2 answers:

krek1111 [17]3 years ago

4 0

27 = 9^(3/2)
.
so x = 4/3

Westkost [7]3 years ago

4 0

Answer:

The value of x is, $\frac{3}{4}$

Explanation:

Given:

The exponential equation $9^{2x} =27$ ....[1]

Exponential Identities:

$(x^a) =x^{ab}$

we can write equation [1] as;

$(3^2)^{2x} = 3^3$ [ $3 \times 3= 9 , 3 \times 3 \times 3 = 27$ we can rewrite as $3^2 =9 , 3^3 =27$ ]

by using above identities, we have

$(3)^{4x} = 3^3$ then,

$4x =3$ [ if $x^a =x^b$ then a =b]

Simplify:

$x = \frac{3}{4}$

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The function C(h)=(2h^2+5h)/(h^3+8) models the concentration of medication in the bloodstream (as a percent) h hours after its i

salantis [7]

Answer:

h ≥ 0
C = 0; concentration eventually decays to nothing
(0, 0) is the only intercept in the domain. It means the concentration in the bloodstream is zero at the time the drug is injected.
1.95 hours

Step-by-step explanation:

1. a. The domain of the function in the context of this problem is h ≥ 0. The maximum value of h will correspond to the time at which the concentration is considered to be negligible. If that time is when the concentration decays to 1% of its peak value, then perhaps the suitable domain is 0 ≤ h ≤ 180.

b. The equation of the vertical asymptote of this function is where the denominator of C(h) is zero, that is ...

h^3 +8 = 0

h = ∛(-8)

h = -2 . . . . the equation of the vertical asymptote

This is not a concern for a medical professional because it is outside the domain of the function.

2. As h gets large the value of the function approaches 2/h, which is to say the horizontal asymptote is C = 0. It means the concentration of medication in the blood eventually decays to zero.

3. C(0) = 0 is the only intercept in the domain of the function. (h, C(h)) = (0, 0)

In the context of this problem, it means the blood concentration of medication is zero at the time of the injection.

4. About 1.95 hours after injection, a maximum concentration of the drug occur in the bloodstream. This value is easily found using a graphing calculator.

Taking the derivative of the function gives you ...

C'(h) = -2(h^4 +5h^3 -16h -20)/(h^3 +8)^2

This has two real zeros and two complex zeros. The positive real zero is near h = 1.94527, about 1.95.

That is, the concentration in the bloodstream reaches a maximum about 1.95 hours after injection into the muscle.