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\(\quad\displaystyle2\log_53 - 3\log_52\\ = \displaystyle\log_5(3^2)-\log_5(2^3)\\ = \displaystyle \log_5(\dfrac{9}{8})\)

\(\boxed {\displaystyle\color{red}a\log b = \log b^a}\\ \color{aqua}and \\\boxed{\displaystyle\color{red}\log a - \log b = \log \dfrac{a}{b}}\)

4¹ = 4

4² = 16

4³ = 64

Therefore 4²ⁿ = XXXXXXXX6     4²ⁿ⁺¹ = XXXXXXXX4

Therefore unit digit of 2354¹⁰⁴⁸ = unit digit of 4¹⁰⁴⁸  = 6.

x # y  =  x² / y

     --->     a # 3  =  a² / 3

If  a # 3  =  9     --->     a² / 3  =  9     --->      a²  =  27     --->     a = 3·sqrt(3)     or     a = - 3·sqrt(3)

You have:  (x^1/3)⁴  =  16

Then you took the fourth root of both sides:

      this will give you either  x^1/3  =  2    or    x^1/3  =  -2

Now you need to cube both sides:

      (x^1/3)³  =  (2)³          and         (x^1/3)³  =  (-2)³    

              x  =  8                                    x  =  -8

But, unless you are using complex numbers, logs don't have negative numbers for bases, so the only answer is:  x = 8.

You started correctly, just keep going!

Let   g(x)  =  x⁴ - 5x³ + 2x² + 7x - 11   with roots p, q, r, and s.

Since the roots are p, q, r, and s:   g(x)  =  (x - p)(x - q)(x - r)(x - s)

Multiply out  (x - p)(x - q)(x - r)(x - s)  

     =  x⁴ - x³p - x³​q - x³​r - x³​s + x²pq + x²​pr + x²​ps + x²​qr + x²​qs + x²​rs - xpqr - xpqs - xqrs - xprs - xqrs + pqrs

     =  x⁴ - (p + q + r + s)x³ + (pq + pr + ps + qr + qs + rs)x² - (pqr + pqs + qrs + prs)x + pqrs

Since the coefficients of the x-cubed terms are equal:  - (p + q + r + s)  =  - 5   --->   p + q + r + s  =  5

Since the coefficients of the x-squared terms are equal:  (pq + pr + ps + qr + qs + rs)  =  2

Since the coefficients of the x-terms are equal:  - (pqr + pqs + qrs + prs)  =  + 7   --->   pqr + pqs + qrs + prs  =  -7

and:  pqrs  =  -11

(a)  Compute pqr+pqs+prs+qrs  =  -7

(b)  Compute 1/p+1/q+1/r+1/s:  

       Write with the common denominator of pqrs:  (qrs + prs + pqs + pqr) / pqrs   =   -7/-11  =  7/11

(c)  Compute p^2+q^2+r^2+s^2:

          (p + q + r + s)²  =  p² + q² + r² + s² + 2pq + 2pr + 2ps + 2qr + 2qs + 2rs

                                   =  p² + q² + r² + s² + 2(pq + pr + ps + 2qr + 2qs + 2rs)

                                   =  p² + q² + r² + s² + 2(2)

                                   =  p² + q² + r² + s² + 4

          But, since  p + q + r + s  =  5

                --->     (5)²  =  p² + q² + r² + s² + 4

                --->      25  =  p² + q² + r² + s² + 4

                --->     p² + q² + r² + s²  =  21

(d)  Compute p^2qrs+pq^2rs+pqr^2s+pqrs^2:

          p^2qrs+pq^2rs+pqr^2s+pqrs^2  =  pqrs(p + q + r + s)  =  -11(p + q + r + s)  =  -11(5)  =  -55

0.545-3[0.545]^1/2

Good evening Guest!

How do i find a formula for the inverse

if this is one to one x - 10 /x + 4

y = x - \(\frac{10}{x} \) + 4

Inverse

x = y - \(\frac{10}{y} \) + 4

xy = y² - 10 + 4y

y² - 10 + 4y - xy = 0

y² + (4 - x) y - 10 = 0

Quadratic formula

\(y = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

a = 1   b = (4 - x)    c = - 10

\(y = {x-4 \pm \sqrt{(4-x)^2+40} \over 2}\)  

This is the inverse function.

Greetings asinus :- )

  !

\(\begin{bmatrix} 1 && 2 \\ 3 && 4 \end{bmatrix}\)

LaTeX code: (\begin{bmatrix}1 && 2  

                     3 && 4 \end{bmatrix}) without the ()

In calculator, Matrices can't be calculated but value of the above one (the 1 2 3 4 one) = 2 x 3 - 1 x 4. 

\(\boxed{\color{red}\text{Dividing a number is same as multiplying its reciprocal}}\)

\(\boxed{\color{olive}\text{Division of a fraction: Exchange the denominator and the numerator and change}\\\color{olive}\div\text{into}\times.} \)

\(\boxed{\color{aqua}\text{If there exists a number x, its reciprocal is}\dfrac{1}{x}}\)

\(\quad\dfrac{1}{6}\div \dfrac{1}{3} \div \dfrac{2}{5}\\=\dfrac{1}{6}\times \dfrac{1}{\dfrac{1}{3}}\times\dfrac{1}{\dfrac{2}{5}}\\ =\dfrac{1}{6}\times 3 \times \dfrac{5}{2}\\=\dfrac{3\times 5}{6\times2}\\=\dfrac{5}{4}\\\color{red}OR\\=1\dfrac{1}{4}\)

x + ln x = 1

ln e^x + ln x = 1

ln xe^x = 1

xe^x = e

Only possible answer is x = 1

Check:

LHS = x + ln x = 1 + ln 1 = 1 + 0 = 1 = RHS.

And please answer there in the original post.

×+ln×=1

x=1

Let p(x) be a polynomial of the form p(x)=2x^3+ax^2+38x+c, such that 4 and 6 are roots of p(x).

(a) What is the third root? 

let the roots be 4,6 and L

\( 4L+6L+4*6=\frac{38}{2}\\ 10L+24=19\\ 10L=-5\\ L=-0.5 \)

The third root is  -0.5

(b) Compute ac.

\(-0.5*4*6=\frac{-c}{2}\\ -24=-c\\ c=24\\ 4+6+-0.5=\frac{-a}{2}\\ 19=-a\\ a=-19\\~\\ ac=-19*24=-456\\~\\ \)

I have not checked my answers Debae, I will leave that task to you.  :)

Hi Dabae,

Let p(x)=1x^3-5x^2+12x-19 have roots a, b, and c. Find the value of 1/(ab)+1/(bc)+1/(ca).

\(a+b+c=--5/1=+5\\ ab+ac+bc=+12/1=+12\\ abc=--19/1=+19\\~\\ \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\frac{c}{abc}+\frac{a}{abc}+\frac{b}{abc}=\frac{a+b+c}{abc}=\frac{5}{19}\)

if the coefficient of x in the expansion of   \(\left(x^2+\frac{k}{x}\right)^5=270, \)

find the value of k.

Using Pascal's triangle the coefficients of the binomial expansion will be 1,5,10,10,5,1

so we have

\(LHS\\ =\left(x^2+\frac{k}{x}\right)^5\\ 1*(x^2)^5*\left(\frac{k}{x}\right)^0\;\;\; +\;\;\;5*(x^2)^4*\left(\frac{k}{x}\right)^1\;\;\; +\;\;\;10*(x^2)^3*\left(\frac{k}{x}\right)^2\;\;\;\\ +\;\;\;10*(x^2)^2*\left(\frac{k}{x}\right)^3\;\;\; +\;\;\;5*(x^2)^1*\left(\frac{k}{x}\right)^4\;\;\; +\;\;\;1*(x^2)^0*\left(\frac{k}{x}\right)^5\\~\\~\\ \mbox{I am only interested in the x term}\\ 10x^4*\left(\frac{k}{x}\right)^3=270x\\ 10k^3=270\\ k^3=27\\ k=3 \)

Formula is: A=h*(a+b)/2 where a and b are the two parallel sides. 

This gives 57=h*(12+(12-5))/2

h=....

10^-9 =nano

How to say "2.8571429e-9

"Two, point, eight, five, seven..............TIMES ten to the MINUS nine"

Since the mean value is 6 the sum of all six scores must be 36 because

(mean=6 x numbers of scores=6)

an equation has appeared!  : 1+2+a+7+b+12=36 ->   a+b=14

We can't solve this one without another equation. 

The median value is in this case the mean value of the two numbers in center (a and 7) 

we know that the median is 5. 

Therfore (a+7)/2=5  --> a+7=10 a=10-7=3 

a=3 gives 3+b=14 in the firat equation. 

b=14-3 

answer: a=3. b=11

1,2,a,7,b,12

median: {1, 2, 3, 7, 11, 12}=5

mean: {1, 2, 3, 7, 11, 12}=6

(1.66*10^-24)(9*10^16)=0.0000001494=1.494 x 10^-7

1/6 divided by 1/3 divided by 2/5=? How do I work this out?

(1/6) / (1/3) / (2/5) = Turn division into multiplication, thus:

1/6 x 3/1 x 5/2 =(1x3x5) / (6x1x2) =15/12 =1 1/4

If repeats are allowed then we have:

26^3 x 10^3 =17,576,000 plates.Note: Some letter combinations may not be allowed such as "swear" words.

If repeats are not allowed, then we have:

26 x 25 x 24 x 10 x 9 x 8 =11,232,000 plates. Note: Same as above.

The first line is contradictory. How can a bunch of positive integers add up and became a negative fraction which is \(-\dfrac{1}{12}\)?

But I will solve this.

\(1+2+... = -\dfrac{1}{12}\\ 1^3+2^3+...... = (1+2+.... )^2=(-\dfrac{1}{12})^2=\dfrac{1}{144}\)

This is still contradictory because the sum of a bunch of third powers of a negative number is a positive number.

You can just type it on your keyboard. Or if you don't like it, it's on the left hand side which in your meaning right hand side.