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2 years ago

Which shows the first step in the solution to the equation log₂x + log₂(x - 6) = 4?

Mathematics

1 answer:

iren [92.7K]2 years ago

6 0

log was used calculate big numbers before calculators

log is a re-arranged way to show a number with an exponent

example

log₂ 16 = 4 means 2^4 = 16

logx(Z) = y means x^y=Z

log(x-6)/log(2) + log(x)/log(2) = 4

(log(x-6)+ log(x))/log(2) = 4

(log(x-6)+ log(x)) = 4log(2)

(log(x-6)x) = log(16)

x=8

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How do you write 1,680 in a scientific notation

AveGali [126]

1,680 (one thousand six hundred eighty) is an even four-digits composite number following 1679 and preceding 1681. In scientific notation, it is written as 1.68 × 103.

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4 years ago

<img src="https://tex.z-dn.net/?f=%20%5Cdisplaystyle%5Crm%5Cint%20%5Climits_%7B0%7D%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D%20%5

umka2103 [35]

Replace $x\mapsto \tan^{-1}(x)$ :

$\displaystyle \int_0^{\frac\pi2} \sqrt[3]{\tan(x)} \ln(\tan(x)) \, dx = \int_0^\infty \frac{\sqrt[3]{x} \ln(x)}{1+x^2} \, dx$

Split the integral at x = 1, and consider the latter one over [1, ∞) in which we replace $x\mapsto\frac1x$ :

$\displaystyle \int_1^\infty \frac{\sqrt[3]{x} \ln(x)}{1+x^2} \, dx = \int_0^1 \frac{\ln\left(\frac1x\right)}{\sqrt[3]{x} \left(1+\frac1{x^2}\right)} \frac{dx}{x^2} = - \int_0^1 \frac{\ln(x)}{\sqrt[3]{x} (1+x^2)} \, dx$

Then the original integral is equivalent to

$\displaystyle \int_0^1 \frac{\ln(x)}{1+x^2} \left(\sqrt[3]{x} - \frac1{\sqrt[3]{x}}\right) \, dx$

Recall that for |x| < 1,

$\displaystyle \sum_{n=0}^\infty x^n = \frac1{1-x}$

so that we can expand the integrand, then interchange the sum and integral to get

$\displaystyle \sum_{n=0}^\infty (-1)^n \int_0^1 \left(x^{2n+\frac13} - x^{2n-\frac13}\right) \ln(x) \, dx$

Integrate by parts, with

$u = \ln(x) \implies du = \dfrac{dx}x$

$du = \left(x^{2n+\frac13} - x^{2n-\frac13}\right) \, dx \implies u = \dfrac{x^{2n+\frac43}}{2n+\frac43} - \dfrac{x^{2n+\frac23}}{2n+\frac23}$

$\implies \displaystyle \sum_{n=0}^\infty (-1)^{n+1} \int_0^1 \left(\dfrac{x^{2n+\frac43}}{2n+\frac43} - \dfrac{x^{2n+\frac13}}{2n-\frac13}\right) \, dx \\\\ = \sum_{n=0}^\infty (-1)^{n+1} \left(\frac1{\left(2n+\frac43\right)^2} - \frac1{\left(2n+\frac23\right)^2}\right) \\\\ = \frac94 \sum_{n=0}^\infty (-1)^{n+1} \left(\frac1{(3n+2)^2} - \frac1{(3n+1)^2}\right)$

Recall the Fourier series we used in an earlier question [27217075]; if $f(x)=\left(x-\frac12\right)^2$ where 0 ≤ x ≤ 1 is a periodic function, then

$\displaystyle f(x) = \frac1{12} + \frac1{\pi^2} \sum_{n=1}^\infty \frac{\cos(2\pi n x)}{n^2}$

$\implies \displaystyle f(x) = \frac1{12} + \frac1{\pi^2} \left(\sum_{n=0}^\infty \frac{\cos(2\pi(3n+1)x)}{(3n+1)^2} + \sum_{n=0}^\infty \frac{\cos(2\pi(3n+2)x)}{(3n+2)^2} + \sum_{n=1}^\infty \frac{\cos(2\pi(3n)x)}{(3n)^2}\right)$

$\implies \displaystyle f(x) = \frac1{12} + \frac1{\pi^2} \left(\sum_{n=0}^\infty \frac{\cos(6\pi n x + 2\pi x)}{(3n+1)^2} + \sum_{n=0}^\infty \frac{\cos(6\pi n x + 4\pi x)}{(3n+2)^2} + \sum_{n=1}^\infty \frac{\cos(6\pi n x)}{(3n)^2}\right)$

Evaluate f and its Fourier expansion at x = 1/2 :

$\displaystyle 0 = \frac1{12} + \frac1{\pi^2} \left(\sum_{n=0}^\infty \frac{(-1)^{n+1}}{(3n+1)^2} + \sum_{n=0}^\infty \frac{(-1)^n}{(3n+2)^2} + \sum_{n=1}^\infty \frac{(-1)^n}{(3n)^2}\right)$

$\implies \displaystyle -\frac{\pi^2}{12} - \frac19 \underbrace{\sum_{n=1}^\infty \frac{(-1)^n}{n^2}}_{-\frac{\pi^2}{12}} = - \sum_{n=0}^\infty (-1)^{n+1} \left(\frac1{(3n+2)^2} - \frac1{(3n+1)^2}\right)$

$\implies \displaystyle \sum_{n=0}^\infty (-1)^{n+1} \left(\frac1{(3n+2)^2} - \frac1{(3n+1)^2}\right) = \frac{2\pi^2}{27}$

So, we conclude that

$\displaystyle \int_0^{\frac\pi2} \sqrt[3]{\tan(x)} \ln(\tan(x)) \, dx = \frac94 \times \frac{2\pi^2}{27} = \boxed{\frac{\pi^2}6}$

3 0

3 years ago

Explain how 4/10÷2 and 4/10×1/2 both equal 2/10.

iris [78.8K]

Answer:

Step-by-step explanation:

when you are taking (4/10) and dividing it by 2. you have a fraction over a fraction. if you remember what you do to the top part of the equation you do to the bottom. so we want to get rid of dividing by 2. (multiple by the reciprocal) the reciprocal of 2 is 1/2. that will cancel out the bottom part of the equation but what you do to the bottom you have to do to the top. so, (4/10) * 1/2 is the same as (4/10) divided by 2. hope that makes sense. it's much easier to write out then type math...

5 0

3 years ago

What is the value of f(x)=23.2x+2.5 when x =-4.2

a_sh-v [17]

Answer:

− 94.94

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Simplify completely<br> x2 + 4x - 45<br> and find the restrictions on the variable.

malfutka [58]

Answer:

(x-5)/(x+1) and x cannot equal -1 because the denominator can't be zero.

Step-by-step explanation:

Factor the numerator:

(x+9)(x-5)

Factor the denominator:

(x+1)(x+9)

So you have

(x+9)(x-5)/(x+1)(x+9); see that you can cross out the (x+9) in both the numerator and denominator?

You're left with (x-5)/(x+1)

The only number that is restricted is -1, because that would make the denominator zero, which is a no no!!

4 0

4 years ago