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4 years ago

²log(x+2)+²log(x-2)≤²log5

1 answer:

6 0

I believe you meant to write
$log_{2}(x+2) + log_{2}(x-2) \leq log_{2}(5)$ ?
If that's the case I'll solve the one I provided but I'll drop the base 2 to type it faster but you need to put it always!
Remember: log a + log b = log (a*b)
So log (x+2) + log (x-2) = log [(x+2)*(x-2)] = log (x^2 - 4)
Now back to the inequality:
log (x^2 - 4) <span>≤ log 5
Raise both sides as powers of 2 ( Since it's the base of your log)
Now,
x^2 - 4 </span><span>≤ 5
Add 4 both sides:
x^2 </span>≤ 9
Square root both sides
x ≤ +3 or x ≤ -3
Reject the -3 solution as it makes both (x + 2) and (x - 2) negative and a log can never have a negative value inside its brackets.
So x <span>≤ 3 But can never be less than 2 as well for the same previous reason.
Hope that helped.</span>

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Write the expression in complete factored form. 2n^2(n-2) - a(n-2)

tatuchka [14]

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Step-by-step explanation:

Given the expression 2n²(n-2) - a(n-2), to write this expression in a complete factor form, simply follow the instruction;

Let's assume the original expression has been broken down into that form in question, if we look at both terms, we will see that n-2 in parenthesis is common to both terms, we can therefore factor out n-2 from both terms as shown;

= 2n²(n-2) - a(n-2)

= n-2(2n² - a)

Hence the complete factor form of the expression is (n-2)(2n² - a) because the expression cannot be simplified any further.

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3 years ago

The joint probability density function of X and Y is given by fX,Y (x, y) = ( 6 7 x 2 + xy 2 if 0 < x < 1, 0 < y < 2

fredd [130]

I'm going to assume the joint density function is

$f_{X,Y}(x,y)=\begin{cases}\frac67(x^2+\frac{xy}2\right)&\text{for }0$

a. In order for $f_{X,Y}$ to be a proper probability density function, the integral over its support must be 1.

$\displaystyle\int_0^2\int_0^1\frac67\left(x^2+\frac{xy}2\right)\,\mathrm dx\,\mathrm dy=\frac67\int_0^2\left(\frac13+\frac y4\right)\,\mathrm dy=1$

b. You get the marginal density $f_X$ by integrating the joint density over all possible values of $Y$ :

$f_X(x)=\displaystyle\int_0^2f_{X,Y}(x,y)\,\mathrm dy=\boxed{\begin{cases}\frac67(2x^2+x)&\text{for }0$

c. We have

$P(X>Y)=\displaystyle\int_0^1\int_0^xf_{X,Y}(x,y)\,\mathrm dy\,\mathrm dx=\int_0^1\frac{15}{14}x^3\,\mathrm dx=\boxed{\frac{15}{56}}$

d. We have

$\displaystyle P\left(X$

and by definition of conditional probability,

$P\left(Y>\dfrac12\mid X\frac12\text{ and }X$

$\displaystyle=\dfrac{28}5\int_{1/2}^2\int_0^{1/2}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\boxed{\frac{69}{80}}$

e. We can find the expectation of $X$ using the marginal distribution found earlier.

$E[X]=\displaystyle\int_0^1xf_X(x)\,\mathrm dx=\frac67\int_0^1(2x^2+x)\,\mathrm dx=\boxed{\frac57}$

f. This part is cut off, but if you're supposed to find the expectation of $Y$ , there are several ways to do so.

Compute the marginal density of $Y$ , then directly compute the expected value.

$f_Y(y)=\displaystyle\int_0^1f_{X,Y}(x,y)\,\mathrm dx=\begin{cases}\frac1{14}(4+3y)&\text{for }0$

$\implies E[Y]=\displaystyle\int_0^2yf_Y(y)\,\mathrm dy=\frac87$

Compute the conditional density of $Y$ given $X=x$ , then use the law of total expectation.

$f_{Y\mid X}(y\mid x)=\dfrac{f_{X,Y}(x,y)}{f_X(x)}=\begin{cases}\frac{2x+y}{4x+2}&\text{for }0$

The law of total expectation says

$E[Y]=E[E[Y\mid X]]$

We have

$E[Y\mid X=x]=\displaystyle\int_0^2yf_{Y\mid X}(y\mid x)\,\mathrm dy=\frac{6x+4}{6x+3}=1+\frac1{6x+3}$

$\implies E[Y\mid X]=1+\dfrac1{6X+3}$

This random variable is undefined only when $X=-\frac12$ which is outside the support of $f_X$ , so we have

$E[Y]=E\left[1+\dfrac1{6X+3}\right]=\displaystyle\int_0^1\left(1+\frac1{6x+3}\right)f_X(x)\,\mathrm dx=\frac87$

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3 years ago