All categories

3 years ago

Prove that x=2 if 4^x=8x...where ^ stands for "to the power of".

1 answer:

4 0

In order to prove this, you simply plug in the number 2 everywhere you see X:

4^(2) = 8(2).

Simplify

16 = 16 √

Since this checks out, x is proven to be 2.

You might be interested in

X+56<533. What is the solution of the inequality?

swat32

X<477

hope this helps<3

7 0

2 years ago

HELP QUICK! 15 POINTS

almond37 [142]

it will be 8.25 hope this helped

6 0

2 years ago

Stacy and Travis are rock climbing. Stacy's rope is 4 feet shorter than 3 times the length of Travis' rope. Stacy's rope is 23 f

abruzzese [7]

First, take off 4 feet of Stacy's rope, which is 19. (23-4=19). Then, divide 19 by 3. It should equal 6.3...

7 0

3 years ago

Find the area of the region that lies inside the first curve and outside the second curve. r = 6 − 6 sin θ, r = 6

yan [13]

Each curve completes one loop over the interval $0\le t\le2\pi$ . Find the intersections of the curves within this interval.

$6-6\sin\theta=6\implies 1-\sin\theta=1\implies \sin\theta=0\implies \theta=0,\theta=\pi$

The region of interest has an area given by the double integral

$\displaystyle\int_\pi^{2\pi}\int_6^{6-6\sin\theta}r\,\mathrm dr\,\mathrm d\theta$

equivalent to the single integral

$\displaystyle\frac12\int_\pi^{2\pi}\bigg((6-6\sin\theta)^2-6^2\bigg)\,\mathrm d\theta$

which evaluates to $9\pi+72$ .

8 0

2 years ago

Find the indefinite integral using the substitution provided.

Nady [450]

Answer: $7\text{Ln}\left(e^{2x}+10\right)+C$

This is the same as writing 7*Ln( e^(2x) + 10) + C

=======================================================

Explanation:

Start with the equation $u = e^{2x}+10$

Apply the derivative and multiply both sides by 7 like so

$u = e^{2x}+10\\\\\frac{du}{dx} = 2e^{2x}\\\\7\frac{du}{dx} = 7*2e^{2x}\\\\7\frac{du}{dx} = 14e^{2x}\\\\7du = 14e^{2x}dx\\\\$

The "multiply both sides by 7" operation was done to turn the 2e^(2x) into 14e^(2x)

This way we can do the following substitutions:

$\displaystyle \int \frac{14e^{2x}}{e^{2x}+10}dx\\\\\\\displaystyle \int \frac{1}{e^{2x}+10}14e^{2x}dx\\\\\\\displaystyle \int \frac{1}{u}7du\\\\\\\displaystyle 7\int \frac{1}{u}du\\\\\\$

Integrating leads to

$\displaystyle 7\int \frac{1}{u}du\\\\\\7\text{Ln}\left(u\right)+C\\\\\\7\text{Ln}\left(e^{2x}+10\right)+C\\\\\\$

Be sure to replace 'u' with e^(2x)+10 since it's likely your teacher wants a function in terms of x. Also, do not forget to have the plus C at the end. This is a common mistake many students forget to do.

To verify the answer, you can apply the derivative to it and you should get back to the original integrand of $\frac{14e^{2x}}{e^{2x}+10}$

4 0

1 year ago