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3 years ago

What is the value of x if e^3x+6 =8? Answer:

Mathematics

2 answers:

geniusboy [140]3 years ago

8 0

Answer:

x = 1/3 ln(2) or approximately 0.23104

Step-by-step explanation:

e^3x+6 =8

Subtract 6 from each side

e^3x+6-6 =8-6

e^3x =2

Take the natural log of each side

ln( e^3x) =ln(2)

3x = ln(2)

divide by 3

3x/3 = 1/3 ln(2)

x = 1/3 ln(2)

x is approximately 0.23104

charle [14.2K]3 years ago

3 0

Answer:

$\displaystyle x = \frac{ln2}{3}$

General Formulas and Concepts:

Pre-Algebra

Equality Properties

Algebra II

Natural logarithms ln and Euler's number e
Logarithmic Property [Exponential]: $\displaystyle log(a^b) = b \cdot log(a)$
Solving logarithmic equations

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle e^{3x} + 6 = 8$

Step 2: Solve for x

[Equality Property] Isolate x term: $\displaystyle e^{3x} = 2$
[Equality Property] ln both sides: $\displaystyle lne^{3x} = ln2$
Rewrite [Logarithmic Property - Exponential]: $\displaystyle 3xlne = ln2$
Simplify: $\displaystyle 3x = ln2$
[Equality Property] Isolate x: $\displaystyle x = \frac{ln2}{3}$

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In the town where Richard lives, peaches are in season from June through

ollegr [7]

Answer:

Richard had bake an Apple pie.

Step-by-step explanation:

Given:

Seasoned fruit From June to August = peaches

Seasoned fruit from September to November = Apples

Also Given:

Richard picked some fruit and baked a pie in the weekend of October.

We need to find what kind of pie he baked.

Now we know peaches are growing from June to august and apples are growing from September to November.

Also Richard picked a fruit on the weekend of October and baked a pie.

Since October lies between September and November and the fruit growing this season is apple.

we can say that;

Richard had bake an Apple pie.

7 0

2 years ago

1 459/500 as a decimal

Ronch [10]

Answer:

1.918

Step-by-step explanation:

1 then do 459/500 x 2 to make 918/1000 then add and convert

6 0

2 years ago

A found raising meeting raised sh 8100 for buying desk for three schools, A,B and C, the money was shared to the schools in the

Gelneren [198K]

Answer:

A: 1/3, 2700
B: 2/9, 1800
C: 4/9, 3600

Step-by-step explanation:

The total number of ratio units is 3+2+4 = 9, so the fractions each school got were ...

A : B : C = 3/9 : 2/9 : 4/9

These fractions multiply the total amount raised:

A got 1/3 = 2700, B got 2/9 = 1800, C got 4/9 = 3600

8 0

3 years ago

A cell phone company charges a monthly rate of $12.95 and $0.25 a minute per call. The bill for m minutes is $21.20. Write an eq

aleksklad [387]

Answer:

y= 12.95x + 0.25

Step-by-step explanation:

4 0

3 years ago

Need your help now please; the assignment is due tonight and this is the last problem I am having trouble with. The red box is t

Kobotan [32]

The volume of the region R bounded by the x-axis is: $\mathbf{\iint_R(x^2+y^2)dA = \int ^{tan^{-1}(4)}_{0} \int^{\frac{2}{cos \theta}}_{0} \ r^3 dr d\theta}$

<h3>What is the volume of the solid (R) on the X-axis?</h3>

If the axis of revolution is the boundary of the plane region and the cross-sections are parallel to the line of revolution, we may use the polar coordinate approach to calculate the volume of the solid.

From the given graph:

The given straight line passes through two points (0,0) and (2,8). Thus, the equation of the straight line becomes:

$\mathbf{y-y_1 = \dfrac{y_2-y_1}{x_2-x_1}(x-x_1)}$

here:

(x₁, y₁) and (x₂, y₂) are two points on the straight line

Suppose we assign (x₁, y₁) = (0, 0) and (x₂, y₂) = (2, 8) from the graph, we have:

$\mathbf{y-0 = \dfrac{8-0}{2-0}(x-0)}$

y = 4x

Now, our region bounded by the three lines are:

y = 0
x = 2
y = 4x

Similarly, the change in polar coordinates is:

x = rcosθ,
y = rsinθ

where;

x² + y² = r² and dA = rdrdθ

Therefore;

rsinθ = 0 i.e. r = 0 or θ = 0
rcosθ = 2 i.e. r = 2/cosθ
rsinθ = 4(rcosθ) ⇒ tan θ = 4; θ = tan⁻¹ (4)

⇒ r = 0 to r = 2/cosθ
θ = 0 to θ = tan⁻¹ (4)

Then:

$\mathbf{\iint_R(x^2+y^2)dA = \int ^{tan^{-1}(4)}_{0} \int^{\frac{2}{cos \theta}}_{0} \ r^2 (rdr d\theta )}$

$\mathbf{\iint_R(x^2+y^2)dA = \int ^{tan^{-1}(4)}_{0} \int^{\frac{2}{cos \theta}}_{0} \ r^3 dr d\theta}$

Learn more about the determining the volume of solids bounded by region R here:

brainly.com/question/14393123

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3 0

1 year ago