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## What is the value of log 43?

George C.
Featured 3 months ago

$\log 43 \approx 1.633$

#### Explanation:

Suppose you know that:

$\log 2 \approx 0.30103$

$\log 3 \approx 0.47712$

Then note that:

$43 = \frac{129}{3} \approx \frac{128}{3} = {2}^{7} / 3$

So

$\log 43 \approx \log \left({2}^{7} / 3\right) = 7 \log 2 - \log 3 \approx 7 \cdot 0.30103 - 0.47712 = 1.63009$

We know that the error is approximately:

$\log \left(\frac{129}{128}\right) = \log 1.0078125 = \frac{\ln 1.0078125}{\ln 10} \approx \frac{0.0078}{2.3} = 0.0034$

So we can confidently give the approximation:

$\log 43 \approx 1.633$

A calculator tells me:

$\log 43 \approx 1.63346845558$

## What is the sum of an infinite geometric series with a first term of 6 and a common ratio of 1/2?

George C.
Featured 3 months ago

$12$

#### Explanation:

The general term of a geometric series can be represented by the formula:

${a}_{n} = a {r}^{n - 1}$

where $a$ is the initial term and $r$ the common ratio.

We find:

$\left(1 - r\right) {\sum}_{n = 1}^{N} a {r}^{n - 1} = {\sum}_{n = 1}^{N} a {r}^{n - 1} - r {\sum}_{n = 1}^{N} a {r}^{n - 1}$

$\textcolor{w h i t e}{\left(1 - r\right) {\sum}_{n = 1}^{N} a {r}^{n - 1}} = {\sum}_{n = 1}^{N} a {r}^{n - 1} - {\sum}_{n = 2}^{N + 1} a {r}^{n - 1}$

$\textcolor{w h i t e}{\left(1 - r\right) {\sum}_{n = 1}^{N} a {r}^{n - 1}} = a + \textcolor{red}{\cancel{\textcolor{b l a c k}{{\sum}_{n = 2}^{N} a {r}^{n - 1}}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{{\sum}_{n = 2}^{N} a {r}^{n - 1}}}} - a {r}^{N}$

$\textcolor{w h i t e}{\left(1 - r\right) {\sum}_{n = 1}^{N} a {r}^{n - 1}} = a \left(1 - {r}^{N}\right)$

Dividing both ends by $\left(1 - r\right)$ we find:

${\sum}_{n = 1}^{N} a {r}^{n - 1} = \frac{a \left(1 - {r}^{N}\right)}{1 - r}$

If $\left\mid r \right\mid < 1$ then ${\lim}_{N \to \infty} {r}^{N} = 0$ and we find:

${\sum}_{n = 1}^{\infty} a {r}^{n - 1} = {\lim}_{N \to \infty} \frac{a \left(1 - {r}^{N}\right)}{1 - r} = \frac{a}{1 - r}$

In the given example, $a = 6$ and $r = \frac{1}{2}$. So $\left\mid r \right\mid < 1$ and:

${\sum}_{n = 1}^{\infty} 6 \cdot {\left(\frac{1}{2}\right)}^{n - 1} = \frac{6}{1 - \frac{1}{2}} = 12$

## How do you use the remainder theorem and synthetic division to find the remainder when 2x^3-7x^2 div x-5?

Featured 2 months ago

The remainder is $= 75$

#### Explanation:

The remainder theorem states that when a polynomial $f \left(x\right)$ is divided by $x - c$

$f \left(x\right) = \left(x - c\right) q \left(x\right) + r \left(x\right)$

$f \left(c\right) = 0 + r$

Here,

$f \left(x\right) = 2 {x}^{3} - 7 {x}^{2}$

and $\left(x - 5\right)$

$f \left(5\right) = 2 \cdot 125 - 175 = 250 - 175 = 75$

The remainder is $= 75$

We now perform the synthetic division

$\textcolor{w h i t e}{a a a a}$$5$$\textcolor{w h i t e}{a a a a}$$|$$\textcolor{w h i t e}{a a a a}$$2$$\textcolor{w h i t e}{a a a a}$$- 7$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a a}$$0$

$\textcolor{w h i t e}{a a a a}$$\textcolor{w h i t e}{a a a a a}$$|$$\textcolor{w h i t e}{a a a a a}$$\textcolor{w h i t e}{a a a a}$$10$$\textcolor{w h i t e}{a a a a}$$15$$\textcolor{w h i t e}{a a a a}$$75$

$\textcolor{w h i t e}{a a a a a a a a a a}$------------------------------------------------------------

$\textcolor{w h i t e}{a a a a}$$\textcolor{w h i t e}{a a a a a}$$\textcolor{w h i t e}{a a a a a a}$$2$$\textcolor{w h i t e}{a a a a}$$3$$\textcolor{w h i t e}{a a a a}$$15$$\textcolor{w h i t e}{a a a a}$$\textcolor{red}{75}$

The remainder is also $= 75$

The quotient is $= 2 {x}^{2} + 3 x + 15$

## How do you find the vertex, focus and directrix of 2y^2+8=x+8y?

Featured 2 months ago

The vertex is $V = \left(0 , 2\right)$
The focus is $F = \left(\frac{1}{8} , 2\right)$
The directrix is $x = - \frac{1}{8}$

#### Explanation:

Let's rearrange the equation by completing the squares

$2 {y}^{2} + 8 = x + 8 y$

$2 {y}^{2} - 8 y = x - 8$

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

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

$2 {\left(y - 2\right)}^{2} = x$

${\left(y - 2\right)}^{2} = \frac{1}{2} x$

This is the equation of a parabola.

We compare this equation to

${\left(y - b\right)}^{2} = 2 p \left(x - a\right)$

$p = \frac{1}{4}$

The vertex is $V = \left(a , b\right) = \left(0 , 2\right)$

The focus is $F = \left(a + \frac{p}{2} , b\right) = \left(\frac{1}{8} , 2\right)$

The directrix is $x = - \frac{1}{8}$

graph{((y-2)^2-x/2)(y-1000(x+1/8))((x-1/8)^2+(y-2)^2-0.001)=0 [-5.222, 5.874, -0.687, 4.86]}

## How to solve this?4^x+9^x+25^x=6^x+10^x+15^x

Ratnaker Mehta
Featured 1 month ago

$x = 0.$

#### Explanation:

Let ${2}^{x} = a , {3}^{x} = b , \mathmr{and} , {5}^{x} = c .$

$\therefore {4}^{x} + {9}^{x} + {25}^{x} = {a}^{2} + {b}^{2} + {c}^{2.} \ldots \ldots \ldots \ldots . \left(1\right) .$

Also, 6^x={(2)(3)}^x=2^x3^x=ab, &, ||ly, 10^x=ca, 15^x=bc...(2)

Hence, using (1), & (2),the given eqn. becomes,

${a}^{2} + {b}^{2} + {c}^{2} - a b - b c - c a = 0 , \mathmr{and} ,$

$\frac{1}{2} \left\{{\left(a - b\right)}^{2} + {\left(b - c\right)}^{2} + {\left(c - a\right)}^{2}\right\} = 0.$

$\Rightarrow a = b = c , \mathmr{and} , {2}^{x} = {3}^{x} = {5}^{x} .$

${2}^{x} = {3}^{x} \Rightarrow {\left(\frac{2}{3}\right)}^{x} = 1 = {\left(\frac{2}{3}\right)}^{0}$

$x = 0.$

${3}^{x} = {5}^{x} \text{ also } \Rightarrow x = 0.$

$\therefore x = 0$ is the Soln.

Enjoy Maths.!

## If Z=x+yi and |(z-3i)/(z+3i)|=1, then z in Argand's diagram lies...?

Ratnaker Mehta
Featured 6 days ago

$z$ lies on the $X -$ Axis or the Real Axis.

#### Explanation:

$| \frac{z - 3 i}{z + 3 i} | = 1 \Rightarrow | z - 3 i | = | z + 3 i |$

$\therefore | x + y i - 3 i | = | x + y i + 3 i | , \ldots \ldots \left[\because , z = x + y i\right]$

$\therefore | x + i \left(y - 3\right) | = | x + i \left(y + 3\right) |$

$\therefore | x + i \left(y - 3\right) {|}^{2} = | x + i \left(y + 3\right) {|}^{2}$

$\therefore {x}^{2} + {\left(y - 3\right)}^{2} = {x}^{2} + {\left(y + 3\right)}^{2} , \ldots . \left[\because , | a + b i {|}^{2} = {a}^{2} + {b}^{2}\right]$

$\therefore - 6 y = 6 y , \mathmr{and} , y = 0.$

This meas that, under the given condition,

$\forall z = x + i y \in {\mathbb{R}}^{2} , y = 0.$

Hence, all such $z$ lies on the $X -$ Axis, or, the Real Axis in

the Argand's Diagram.

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