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## How do you solve 6^x=13?

Jim G.
Featured 2 months ago

x≈1.432" to 3 decimal places"

#### Explanation:

Using the $\textcolor{b l u e}{\text{law of logarithms}}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{\log {a}^{n} = n \log a} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

${6}^{x} = 13$

Take the ln ( natural log ) of both sides.

$\Rightarrow \ln {6}^{x} = \ln 13$

$\Rightarrow x \ln 6 = \ln 13$

rArrx=ln13/ln6≈1.432" to 3 decimal places"

## How do you find the zeros with multiplicity for the function p(x)=(x^3-8)(x^5-4x^3)?

George C.
Featured 2 months ago

$p \left(x\right)$ has zeros:

$0$ with multiplicity $3$

$2$ with multiplicity $2$

$- 2$ with multiplicity $1$

$- 1 + \sqrt{3} i$ with multiplicity $1$

$- 1 - \sqrt{3} i$ with multiplicity $1$

#### Explanation:

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

The difference of cubes identity can be written:

${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

We find:

$p \left(x\right) = \left({x}^{3} - 8\right) \left({x}^{5} - 4 {x}^{3}\right)$

$\textcolor{w h i t e}{p \left(x\right)} = \left({x}^{3} - {2}^{3}\right) {x}^{3} \left({x}^{2} - {2}^{2}\right)$

$\textcolor{w h i t e}{p \left(x\right)} = \left(x - 2\right) \left({x}^{2} + 2 x + 4\right) {x}^{3} \left(x - 2\right) \left(x + 2\right)$

$\textcolor{w h i t e}{p \left(x\right)} = {x}^{3} {\left(x - 2\right)}^{2} \left(x + 2\right) \left({x}^{2} + 2 x + 4\right)$

$\textcolor{w h i t e}{p \left(x\right)} = {x}^{3} {\left(x - 2\right)}^{2} \left(x + 2\right) \left({x}^{2} + 2 x + 1 + 3\right)$

$\textcolor{w h i t e}{p \left(x\right)} = {x}^{3} {\left(x - 2\right)}^{2} \left(x + 2\right) \left({\left(x + 1\right)}^{2} - {\left(\sqrt{3} i\right)}^{2}\right)$

$\textcolor{w h i t e}{p \left(x\right)} = {x}^{3} {\left(x - 2\right)}^{2} \left(x + 2\right) \left(\left(x + 1\right) - \sqrt{3} i\right) \left(\left(x + 1\right) + \sqrt{3} i\right)$

$\textcolor{w h i t e}{p \left(x\right)} = {x}^{3} {\left(x - 2\right)}^{2} \left(x + 2\right) \left(x + 1 - \sqrt{3} i\right) \left(x + 1 + \sqrt{3} i\right)$

Hence zeros:

$0$ with multiplicity $3$

$2$ with multiplicity $2$

$- 2$ with multiplicity $1$

$- 1 + \sqrt{3} i$ with multiplicity $1$

$- 1 - \sqrt{3} i$ with multiplicity $1$

## How do you graph y=(3x^3+1)/(4x^2-32) using asymptotes, intercepts, end behavior?

Featured 2 months ago

The vertical asymptotes are $x = \sqrt{8}$ and $x = - \sqrt{8}$
The slant asymptote is $y = \frac{3}{4} x$
No horozontal asymptote.

#### Explanation:

Let's factorise the denominator $\left(4 {x}^{2} - 32\right)$

$= 4 \left({x}^{2} - 8\right) = 4 \left(x + \sqrt{8}\right) \left(x - \sqrt{8}\right)$

The domain of $y$ is ${D}_{y} = \mathbb{R} - \left\{\sqrt{8} , - \sqrt{8}\right\}$

As we cannot divide by $0$,

So $x \ne \sqrt{8}$ and $x \ne - \sqrt{8}$

The vertical asymptotes are $x = \sqrt{8}$ and $x = - \sqrt{8}$

As the degree of the numerator is $>$ the degree of the denominator, we expect a slant asymptote.

Let's do a long division

$\textcolor{w h i t e}{a a a a}$$3 {x}^{3} + 1$$\textcolor{w h i t e}{a a a a}$∣$4 {x}^{2} - 32$

$\textcolor{w h i t e}{a a a a}$$3 {x}^{3} - 24 x$$\textcolor{w h i t e}{a a}$∣$\frac{3 x}{4}$

$\textcolor{w h i t e}{a a a a a}$$0 - 24 x + 1$

So, $\frac{3 {x}^{3} + 1}{4 {x}^{2} - 32} = \frac{3 x}{4} - \frac{24 x + 1}{4 {x}^{2} + 32}$

The slant asymptote is $y = \frac{3}{4} x$

To calculate the limits, we use the terms of highest degree.

${\lim}_{x \to + \infty} y = {\lim}_{x \to + \infty} \frac{3 {x}^{3}}{4 {x}^{2}} = {\lim}_{x \to + \infty} \frac{3 x}{4} = + \infty$

${\lim}_{x \to - \infty} y = {\lim}_{x \to - \infty} \frac{3 {x}^{3}}{4 {x}^{2}} = {\lim}_{x \to - \infty} \frac{3 x}{4} = - \infty$

There are no horizontal asymptote

${\lim}_{x \to - {\sqrt{8}}^{-}} = - \infty$

${\lim}_{x \to - {\sqrt{8}}^{+}} = + \infty$

${\lim}_{x \to {\sqrt{8}}^{-}} = - \infty$

${\lim}_{x \to {\sqrt{8}}^{+}} = + \infty$

When $x = 0$, $\implies$, $y = - \frac{1}{32}$

When $y = 0$, $0 >$, $x = {\left(- \frac{1}{3}\right)}^{\frac{1}{3}}$

graph{(y-(3x^3+1)/(4x^2-32))(y-x3/4)=0 [-28.86, 28.9, -14.43, 14.43]}

## How do you simplify (2+5i)/(5+4i)?

Featured 1 month ago

The answer is $= \frac{30}{41} + \frac{17 i}{41}$

#### Explanation:

If you want to simplify a quotient of complex numbers , multiply numerator and denominator by the conjugate of the denominator.

$z = {z}_{1} / {z}_{2} = \frac{{z}_{1} {\overline{z}}_{2}}{{z}_{2} {\overline{z}}_{2}}$

The conjugate of $\left(a + i b\right)$ is $\left(a - i b\right)$

And ${i}^{2} = - 1$

Here, ${z}_{1} = 2 + 5 i$ and ${z}_{2} = 5 + 4 i$

So $\frac{2 + 5 i}{5 + 4 i} = \frac{\left(2 + 5 i\right) \left(5 - 4 i\right)}{\left(5 + 4 i\right) \left(5 - 4 i\right)}$

$= \frac{10 - 8 i + 25 i - 20 {i}^{2}}{25 - 16 {i}^{2}}$

$= \frac{30 + 17 i}{41}$

$= \frac{30}{41} + \frac{17 i}{41}$

## What is the quotient (3x^4-4x^2+ 8x-1)div (x-2)?

Morgan
Featured 4 weeks ago

$3 {x}^{3} + 6 {x}^{2} + 8 x + 24 + \frac{47}{x - 2}$

#### Explanation:

We set up the long division of a polynomial by a simple monomial like this:

(x-2))bar(3x^4-4x^2+8x-1)

It works just like the long (numerical) division most of us learned back in elementary school, except now we're dividing with variables.

First we check: how many times does our leading $x$ term in the divisor, in this case just $x$ (coefficient of $1$), go into our leading $x$ term in the dividend, $3 {x}^{4}$? We would have to multiply $x$ by $3 {x}^{3}$ to get $3 {x}^{4}$. Just like in long division with integers, we put this above the bar.

$\textcolor{w h i t e}{S P A C E} 3 {x}^{3}$
(x-2))bar(3x^4-4x^2+8x-1)

Now, we multiply $3 {x}^{3}$ by the divisor and subtract that from $3 {x}^{4}$.

$\textcolor{w h i t e}{S P A C E} 3 {x}^{3}$
(x-2))bar(3x^4-4x^2+8x-1)
$\textcolor{w h i t e}{S P A} - 3 {x}^{3} \left(x - 2\right)$

$\implies$

$\textcolor{w h i t e}{S P A C E} 3 {x}^{3}$
(x-2))bar(3x^4-4x^2+8x-1)
$\textcolor{w h i t e}{S P A} - 3 {x}^{4} + 6 {x}^{3}$

We can subtract $3 {x}^{4}$ from $3 {x}^{4}$ to get $0$. We see that we have no ${x}^{3}$ term in the dividend, so we treat this as though we're adding $6 {x}^{3}$ to $0$.

Now, we check: how many times does our leading coefficient $x$ in the divisor go into $6 {x}^{3}$? We would have to multiply $x$ by $6 {x}^{2}$ to get $6 {x}^{3}$. Similarly to above:

$\implies$

$\textcolor{w h i t e}{S P A C E} 3 {x}^{3} + 6 {x}^{2}$
(x-2))bar(3x^4-4x^2+8x-1)
$\textcolor{w h i t e}{S P A} - 3 {x}^{4} + 6 {x}^{3}$
$\textcolor{w h i t e}{S P A C E S P A} 6 {x}^{3}$
$\textcolor{w h i t e}{S P A C E} - 6 {x}^{2} \left(x - 2\right)$

$\implies$

$\textcolor{w h i t e}{S P A C E} 3 {x}^{3} + 6 {x}^{2}$
(x-2))bar(3x^4-4x^2+8x-1)
$\textcolor{w h i t e}{S P A} - 3 {x}^{4} + 6 {x}^{3}$
$\textcolor{w h i t e}{S P A C E S P A} 6 {x}^{3}$
$\textcolor{w h i t e}{S P A C E S} - 6 {x}^{3} + 12 {x}^{2}$

This gets rid of our ${x}^{3}$ term, but now we have an ${x}^{2}$ term. We also have an ${x}^{2}$ term in the dividend, so we can add this to the current remainder.

$\implies$

$\textcolor{w h i t e}{S P A C E} 3 {x}^{3} + 6 {x}^{2}$
(x-2))bar(3x^4-4x^2+8x-1)
$\textcolor{w h i t e}{S P A} - 3 {x}^{4} + 6 {x}^{3}$
$\textcolor{w h i t e}{S P A C E S P A} 6 {x}^{3}$
$\textcolor{w h i t e}{S P A C E S P A} + 12 {x}^{2}$

$\implies$

$\textcolor{w h i t e}{S P A C E} 3 {x}^{3} + 6 {x}^{2}$
(x-2))bar(3x^4-4x^2+8x-1)
$\textcolor{w h i t e}{S P A} - 3 {x}^{4} + 6 {x}^{3}$
$\textcolor{w h i t e}{S P A C E S P A} 6 {x}^{3}$
$\textcolor{w h i t e}{S P A C E S P A C} 8 {x}^{2}$

Next we check: how many times does our leading coefficient $x$ in the divisor go into $8 {x}^{2}$? We would have to multiply $x$ by $8 x$ to get $8 {x}^{2}$.

$\implies$

$\textcolor{w h i t e}{S P A C E} 3 {x}^{3} + 6 {x}^{2} + 8 x$
(x-2))bar(3x^4-4x^2+8x-1)
$\textcolor{w h i t e}{S P A} - 3 {x}^{4} + 6 {x}^{3}$
$\textcolor{w h i t e}{S P A C E S P A} 6 {x}^{3}$
$\textcolor{w h i t e}{S P A C E S P A C} 8 {x}^{2}$
$\textcolor{w h i t e}{S P A C E S P} - 8 x \left(x - 2\right)$

$\implies$

$\textcolor{w h i t e}{S P A C E} 3 {x}^{3} + 6 {x}^{2} + 8 x$
(x-2))bar(3x^4-4x^2+8x-1)
$\textcolor{w h i t e}{S P A} - 3 {x}^{4} + 6 {x}^{3}$
$\textcolor{w h i t e}{S P A C E S P A} 6 {x}^{3}$
$\textcolor{w h i t e}{S P A C E S P A C} 8 {x}^{2}$
$\textcolor{w h i t e}{S P A C E S P} - 8 {x}^{2} + 16 x$

$\implies$

$\textcolor{w h i t e}{S P A C E} 3 {x}^{3} + 6 {x}^{2} + 8 x$
(x-2))bar(3x^4-4x^2+8x-1)
$\textcolor{w h i t e}{S P A} - 3 {x}^{4} + 6 {x}^{3}$
$\textcolor{w h i t e}{S P A C E S P A} 6 {x}^{3}$
$\textcolor{w h i t e}{S P A C E S P A C} 8 {x}^{2}$
$\textcolor{w h i t e}{S P A C E S P A C E S} + 16 x$

Now we add $16 x$ to the $8 x$ term in the dividend.

$\implies$

$\textcolor{w h i t e}{S P A C E} 3 {x}^{3} + 6 {x}^{2} + 8 x$
(x-2))bar(3x^4-4x^2+8x-1)
$\textcolor{w h i t e}{S P A} - 3 {x}^{4} + 6 {x}^{3}$
$\textcolor{w h i t e}{S P A C E S P A} 6 {x}^{3}$
$\textcolor{w h i t e}{S P A C E S P A C} 8 {x}^{2}$
$\textcolor{w h i t e}{S P A C E S P A C E S P A} 24 x$

Next we check: how many times does our leading coefficient $x$ in the divisor go into $24 x$? We would have to multiply $x$ by $24$ to get $24 x$.

$\implies$

$\textcolor{w h i t e}{S P A C E} 3 {x}^{3} + 6 {x}^{2} + 8 x + 24$
(x-2))bar(3x^4-4x^2+8x-1)
$\textcolor{w h i t e}{S P A} - 3 {x}^{4} + 6 {x}^{3}$
$\textcolor{w h i t e}{S P A C E S P A} 6 {x}^{3}$
$\textcolor{w h i t e}{S P A C E S P A C} 8 {x}^{2}$
$\textcolor{w h i t e}{S P A C E S P A C E S P A} 24 x$
$\textcolor{w h i t e}{S P A C E S P A C E} - 24 \left(x - 2\right)$

$\implies$

$\textcolor{w h i t e}{S P A C E} 3 {x}^{3} + 6 {x}^{2} + 8 x + 24$
(x-2))bar(3x^4-4x^2+8x-1)
$\textcolor{w h i t e}{S P A} - 3 {x}^{4} + 6 {x}^{3}$
$\textcolor{w h i t e}{S P A C E S P A} 6 {x}^{3}$
$\textcolor{w h i t e}{S P A C E S P A C} 8 {x}^{2}$
$\textcolor{w h i t e}{S P A C E S P A C E S P A} 24 x$
$\textcolor{w h i t e}{S P A C E S P A C E} - 24 x + 48$

$\implies$

$\textcolor{w h i t e}{S P A C E} 3 {x}^{3} + 6 {x}^{2} + 8 x + 24$
(x-2))bar(3x^4-4x^2+8x-1)
$\textcolor{w h i t e}{S P A} - 3 {x}^{4} + 6 {x}^{3}$
$\textcolor{w h i t e}{S P A C E S P A} 6 {x}^{3}$
$\textcolor{w h i t e}{S P A C E S P A C} 8 {x}^{2}$
$\textcolor{w h i t e}{S P A C E S P A C E S P A} 24 x$
$\textcolor{w h i t e}{S P A C E S P A C E S P A C E} 48$

Now we add $48$ to the $- 1$ in our dividend.

$\implies$

$\textcolor{w h i t e}{S P A C E} 3 {x}^{3} + 6 {x}^{2} + 8 x + 24$
(x-2))bar(3x^4-4x^2+8x-1)
$\textcolor{w h i t e}{S P A} - 3 {x}^{4} + 6 {x}^{3}$
$\textcolor{w h i t e}{S P A C E S P A} 6 {x}^{3}$
$\textcolor{w h i t e}{S P A C E S P A C} 8 {x}^{2}$
$\textcolor{w h i t e}{S P A C E S P A C E S P A} 24 x$
$\textcolor{w h i t e}{S P A C E S P A C E S P A C E} 47$

And of course, $x$ goes into $47$ zero times. This leaves us with a remainder.

The final answer is therefore $3 {x}^{3} + 6 {x}^{2} + 8 x + 24 + \frac{47}{x - 2}$

## Given -f(x), how do you describe the transformation?

Geoff K.
Featured 3 weeks ago

A reflection about the $x$-axis.

#### Explanation:

For any function $f$, its output at any given input $x$ is just $f \left(x\right)$.

When we take that output $f \left(x\right)$ and make it negative (i.e. $- f \left(x\right)$), we're just flipping the sign of the output—positives become negatives, and vice versa.

Let's say that when $x = 3 ,$ we have $f \left(x\right) = 5$.
Thus, when $x = 3 ,$ we have $\text{-"f(x)="-} 5$.

All that happens is the sign of the output value changes—points that were once above the $x$-axis are now below, with no shift left or right.

This is simply stated as a reflection where the mirror is the $x$-axis, also called a reflection about the $x$-axis.

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