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y  = k /x^2

We need to find k

And we know that

9 =  k / 1      multiply both sides by 1

9  = k

So......when y  =  16, we have

16  = 9 / x^2     rearrange as

x^2  =  9 /16        take the positive root

x  =  3 / 4

If the domain of t lies between -infinity and +infinity, the range of the right hand side lies between 0 and 1.  Hence there is no value of t that allows the right hand side to equal 95350.

Are you SURE you have entered the equation correctly?   Is a negative sign missing somewhere?   Is it perhaps (0.5^.8t) ?  SOmeone might be able to sove this , but it is long!

Divide both sides by 3 to get

2.06667 e^.64t = e^(t+1)      divide by  e^.64t

2.06667 = e ^(t+1) / e^(.64t)     take ln of both sides

.72594 = t+1   - .64t)

-.2740  = . 36 t

t =-.7613

If 

$a=\sqrt4+\sqrt5+a\\ then\\ 0=\sqrt4+\sqrt5\\ 0=2+\sqrt5\\ 2=-\sqrt5$

This is obviously nonsense.

so

This equation has no real solutions.  In fact I do not think it has any solutions at all.

Perhaps you wrote the question incorrrectly?

Find all real x that satisfy (2^x - 4)^3 + (4^x - 2)^3 = (4^x + 2^x - 6)^3

$\begin{array}{|lrcll|} \hline (2^x - 4)^3 + (4^x - 2)^3 = (4^x + 2^x - 6)^3 \\ \qquad\text{We substitute:} ~ a = 2^x,~ b = 4^x,~ c = -4,~ d = -2 \\ (a +c)^3 + (b +d)^3 = (a+b+c+d)^3 \\ \hline \end{array}\\ \small{ \begin{array}{|lrcll|} \hline a^3+3ac(a+c)+c^3+b^3+3bd(b+d)+d^3 & =& a^3 + b^3 + c^3+ d^3 \\ && + 6 a b c + 6 a b d + 6 a c d + 6 b c d \\ & & + 3 a^2 b + 3 a^2 c + 3 a^2 d + 3 a b^2 \\ & & + 3 a c^2 + 3 a d^2 + 3 b^2 c + 3 b^2 d \\ & & + 3 b c^2 + 3 b d^2 + 3 c^2 d + 3 c d^2 \\ \not{a^3}+\not{b^3}+\not{c^3}+\not{d^3}+\not{3a^2c}+\not{3ac^2}+\not{3b^2d}+\not{3bd^2} & =& \not{a^3} + \not{b^3} + \not{c^3}+ \not{d^3} \\ && + 6 a b c + 6 a b d + 6 a c d + 6 b c d \\ & & + 3 a^2 b + \not{3 a^2 c} + 3 a^2 d + 3 a b^2 \\ & & + \not{3 a c^2} + 3 a d^2 + 3 b^2 c + \not{3 b^2 d} \\ & & + 3 b c^2 + \not{3 b d^2} + 3 c^2 d + 3 c d^2 \\ \hline \end{array}}\\ \small{ \begin{array}{|rcll|} \hline 0 &=& 6 a b c + 6 a b d + 6 a c d + 6 b c d + 3 a^2 b + 3 a^2 d + 3 a b^2 + 3 a d^2 + 3 b^2 c + 3 b c^2 + 3 c^2 d + 3 c d^2 ~ | ~: 3 \\ 0 &=& 2 a b c + 2 a b d + 2 a c d + 2 b c d + a^2 b + a^2 d + a b^2 + a d^2 + b^2 c + b c^2 + c^2 d + c d^2 \\ 0 &=& a^2(b+d)+b^2(a+c)+c^2(b+d)+d^2(a+c)+2bd(a+c)+2ac(b+d) \\ 0 &=& (b+d)(a^2+2ac+c^2)+(a+c)(b^2+bd+d^2) \\ 0 &=& (b+d)(a+c)^2+(a+c)(b+d)^2 \\ 0 &=& (b+d)(a+c)(a+b+c+d) \\ \hline \end{array}}$

Solution:

$\begin{array}{|rcll|} \hline \mathbf{b+d} &\mathbf{=}& \mathbf{0} \\ 4^x-2 &=& 0 \\ 4^x &=& 2 \\ 2^{2x} &=& 2^1 \\ 2x &=& 1 \\ \mathbf{x_1} &\mathbf{=}& \mathbf{\dfrac{1}{2}} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \mathbf{a+c} &\mathbf{=}& \mathbf{0} \\ 2^x-4 &=& 0 \\ 2^x &=& 4 \\ 2^{x} &=& 2^2 \\ x &=& 2 \\ \mathbf{x_2} &\mathbf{=}& \mathbf{2} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \mathbf{a+b+c+d} &\mathbf{=}& \mathbf{0} \\ 2^x + 4^x-6 &=& 0 \\ (2^x-2)(2^x+3) &=& 0 \\ \hline 2^x-2 &=& 0 \\ 2^x &=& 2^1 \\ x &=& 1 \\ \mathbf{x_3} &\mathbf{=}& \mathbf{1} \\\\ 2^x+3 &=& 0 \\ 2^x &=& -3 \\ x\log(2) &=& \log(-3) \quad & |\quad \text{ no solution, $log(-3)$ complex } \\ \hline \end{array}$

Here is the graph. From the graph it certainly appears that  3^(x-1) will always be bigger than 2^x-7

Find an ordered triple (x,y,z) of real numbers satisfying x<= y<= z and the system of equations
sqrtx + sqrty +sqrtz = 10
x+y+z=38
(sqrtxyz) = 30

$\begin{array}{|lrcll|} \hline (1) & \sqrt{x} + \sqrt{y} +\sqrt{z} &=& 10 \\ (2) & x+y+z &=& 38 \\ (3) & \sqrt{xyz} &=& 30 \\ \hline \end{array}$

1.

$\begin{array}{|lrcll|} \hline & (\sqrt{x} + \sqrt{y} +\sqrt{z})^2 &=& 10^2 \\ & \underbrace{x+y+z}_{=38} +2( \sqrt{xy}+\sqrt{xz}+\sqrt{yz} ) &=& 100 \\ & 38 +2( \sqrt{xy}+\sqrt{xz}+\sqrt{yz} ) &=& 100 \quad & | \quad -38 \\ & 2( \sqrt{xy}+\sqrt{xz}+\sqrt{yz} ) &=& 62 \quad & | \quad :2 \\ \mathbf{(4)}& \mathbf{\sqrt{xy}+\sqrt{xz}+\sqrt{yz}} &\mathbf{=}& \mathbf{31} \\ \hline \end{array}$

$\begin{array}{lcll} \text{We substitute:} \\ \quad x_1 = \sqrt{x} & \text{ or }& x = x_1^2 \\ \quad x_2 = \sqrt{y} & \text{ or }& y = x_2^2 \\ \quad x_3 = \sqrt{z} & \text{ or }& z = x_3^2 \\ \end{array}$

$\begin{array}{|rcll|} \hline x_1+x_2+x_3 &=& \sqrt{x} + \sqrt{y} +\sqrt{z} \\ &=& 10 \\ x_1\cdot x_2 \cdot x_3 &=& \sqrt{xyz} \\ &=& 30 \\ x_1\cdot x_2 + x_1 \cdot x_3 + x_2 \cdot x_3 &=& \sqrt{xy}+\sqrt{xz}+\sqrt{yz} \\ &=& 31 \\ \hline \end{array}$

$\begin{array}{lcll} \text{We set:} \\ \quad p &=& -(x_1+x_2+x_3) \\ &=& -10 \\ \quad q &=& x_1\cdot x_2 + x_1 \cdot x_3 + x_2 \cdot x_3 \\ &=& 31 \\ \quad r &=& -(x_1\cdot x_2 \cdot x_3) \\ &=& -30 \\ \end{array}$

$\begin{array}{lrcll} \text{The cubic equation:} \\ \boxed{ x^3 - 10x^2+31x-30 = 0 } \\ \end{array}$

The first solution is $x_1 = 2$

Long division:

$\begin{array}{rcll} x^2-8x+15 = (x-3)(x-5) \end{array}$

so $x_2 = 3$ and $x_3 = 5$

$\begin{array}{|rcll|} \hline x_1 = 2 & x &=& x_1^2 \\ & \mathbf{x} &\mathbf{=}& \mathbf{4} \\\\ x_2 = 3 & y &=& x_2^2 \\ & \mathbf{y} &\mathbf{=}& \mathbf{9} \\\\ x_3 = 5 & z &=& x_3^2 \\ & \mathbf{z} &\mathbf{=}& \mathbf{25} \\ \hline \end{array}$

The ordered triple (4, 9, 25) of real numbers satisfying $4\le 9\le 25$ and the system of equations.

Find a polynomial $f(x)$ of degree $5$ such that both of these properties hold:
$f(x)-1$ is divisible by $(x-1)^3$.
$f(x)$ is divisible by $x^3$.

1. $\bf{\text{ $\mathbf{f(x)}$ is divisible by $\mathbf{x^3}~$ ?}}$

$\begin{array}{lcll} \text{Let $f(x) = a(x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)$ is a polynomial of degree $5$.} \\ \text{If f(x) is divisible by $x^3$, than $x_1=x_2=x_3=0 $} \\ \qquad f(x) = a(x-0)(x-0)(x-0)(x-x_4)(x-x_5) \\ \qquad \boxed{f(x) =ax^3(x-x_4)(x-x_5)} \text{ This polynom is divisible by $x^3$} \\ \text{expand to:} \\ \qquad f(x)= ax^5 - a(x_4+x_5)x^4 + ax_4x_5x^3 \qquad \text{Let $A=a$, $~B=a(x_4+x_5)$, and $~C = ax_4x_5 $ } \\ \text{finally we have:} \\ \qquad \boxed{f(x) = Ax^5+Bx^4+Cx^3 \qquad (1) } \text{ This polynom is divisible by $x^3$} \\ \end{array}$

2. $\bf{\text{ $\mathbf{f(x)-1}$ is divisible by $\mathbf{(x-1)^3}~$ ?}}$

$\begin{array}{lcll} \text{Say $P(x) $ is a polynom divisible by $(x-1)^3$ .} \\ \text{Set:}\\ \qquad P(x)=b\cdot(x-1)^3 \\ \qquad \text{The root is $x = 1$ }\\ \text{so}\\ \qquad P(1)=b(1-1)^3=0 \\ \text{Set also:}\\ \qquad \text{ $P'(x)= 3b(x-1)^2 $}\\ \text{so}\\ \qquad P'(1)= 3b(1-1)^2=0 \\ \text{Set finally:}\\ \qquad \text{ $P''(x)= 6b(x-1) $}\\ \text{so}\\ \qquad P''(1)= 6b(1-1)^2=0 \\\\ \text{Conclusion} \\ \text{If $P(x)$ is divisible by $(x-1)^3$, so $\boxed{\mathbf{P(1)=P'(1)=P''(1)=0}\qquad (2)}$ at the root $x=1$ } \\ \end{array}$

$\begin{array}{lrcll} \text{We set:}\\ \text{1)}& f(x) - 1 &=& P(x) \quad & | \quad \text{$f(x)-1$ and $P(x)$ are divisible by $(x-1)^3 $ }\\ & f(1) -1 &=& P(1) = 0 \quad & | \quad \rightarrow (2) \\\\ \text{2)}& (f(x) - 1)' &=& P'(x) \\ & f'(x) &=& P'(x) \\ & f'(1) &=& P'(1) = 0 \quad & | \quad \rightarrow (2) \\\\ \text{3)}& (f(x) - 1)'' &=& P''(x) \\ & f''(x) &=& P''(x) \\ & f''(1) &=& P''(1) = 0 \quad & | \quad \rightarrow (2) \\ \end{array}$

$\begin{array}{lrcll} \text{We see:}\\ & \boxed{ \begin{array}{r} f(1) -1 = 0 \qquad (3) \\ f'(1) = 0 \qquad (4) \\ f''(1)= 0 \qquad (5) \\ \end{array} } \end{array}$

$\begin{array}{lrcll} \text{We calculate:}\\ & f(x) &=& Ax^5+Bx^4+Cx^3 \\ & f(x) -1 &=& Ax^5+Bx^4+Cx^3 - 1 \\ & f(1)-1 &=& A+B+C -1 = 0 \quad & | \quad \rightarrow (3) \\\\ & f'(x) &=& 5Ax^4+4Bx^3+3Cx^2 \\ & f'(1) &=& 5A+4B+3C = 0 \quad & | \quad \rightarrow (4) \\\\ & f''(x) &=& 20Ax^3+12Bx^2+6Cx \\ & f''(1) &=& 20A+12B+6C = 0 \quad & | \quad \rightarrow (5) \\ \end{array}$

$\begin{array}{lrcll} \text{Solve the Simultaneous Equations, we calculate $A~$, $B~$, $C~$ :} \\ \end{array}\\ \begin{array}{lrcll} A+B+C -1 = 0 \quad \Rightarrow & A+B+C &=& 1 \\ & 5A+4B+3C &=& 0 \\ & 20A+12B+6C &=& 0 \\ \end{array}$

Cramer's Rule:

$\begin{array}{rcrcrcr} 1\cdot A &+& 1\cdot B &+& 1\cdot C &=& 1 \\ 5\cdot A &+& 4\cdot B &+& 3\cdot C &=& 0 \\ 20\cdot A &+& 12\cdot B &+& 6\cdot C &=& 0 \\ \end{array}\\ \small{ \begin{array}{lcl} \\ \text{Determinant denominator} &=& \begin{vmatrix} 1&1&1 \\ 5&4&3 \\ 20&12&6 \\ \end{vmatrix}\\ \\ &=& 1\cdot 4\cdot 6 + 5\cdot 12\cdot 1 +20\cdot 1\cdot 3 - 20\cdot 4\cdot 1 -1\cdot 12\cdot 3 -5\cdot 1\cdot 6\\ &=& 24+60+60-80-36-30 \\ &=& -2 \\ \end{array} }$

$\begin{array}{lcl} A &=& \dfrac{ \begin{vmatrix} 1&1&1 \\ 0&4&3 \\ 0&12&6 \\ \end{vmatrix} }{-2}\\\\ &=&\dfrac{ 1\cdot 4\cdot 6 - 1\cdot 12\cdot 3 } {-2}\\ &=&\dfrac{ -12 } {-2}\\\\ \mathbf{A} & \mathbf{=} & \mathbf{6}\\ \end{array} \begin{array}{lcl} B &=& \dfrac{ \begin{vmatrix} 1&1&1 \\ 5&0&3 \\ 20&0&6 \\ \end{vmatrix} }{-2}\\\\ &=&\dfrac{ 20\cdot 1\cdot 3 - 5\cdot 1\cdot 6 } {-2}\\ &=&\dfrac{ 30 } {-2}\\\\ \mathbf{B} & \mathbf{=} & \mathbf{-15}\\ \end{array} \begin{array}{lcl} C &=& \dfrac{ \begin{vmatrix} 1&1&1 \\ 5&4&0 \\ 20&12&0 \\ \end{vmatrix} }{-2}\\\\ &=&\dfrac{ 5\cdot 12\cdot 1 - 20\cdot 4\cdot 1 } {-2}\\ &=&\dfrac{ -20 } {-2}\\\\ \mathbf{C} & \mathbf{=} & \mathbf{10}\\ \end{array}$

$\begin{array}{lrcll} \text{The Polynom $f(x) = Ax^5+Bx^4+Cx^3$ with $A=6$, $~B=-15$, and $~C = 10$}\\ \quad \boxed{f(x)=6x^5-15x^4+10x^3} \end{array}$

Proof:

$\begin{array}{|rcll|} \hline f(3) &=& 6\cdot 3^5-15\cdot 3^4 + 10 \cdot 3^3 \\ &=& 1458-1215+270 \\ &=& 513 \\ \frac{f(3)}{3^3} &=& \frac{513}{3^3} \\ &=& 19\ \checkmark \\ \frac{f(3)-1}{(3-1)^3} &=& \frac{513-1}{2^3} \\ &=& \frac{512}{8} \\ &=& 64\ \checkmark \\ \hline \end{array}$

KillerSteve is correct!!!!.....here's a slightly easier method...

(1/3)log x  =  log 8       and we can write

log x ^(1/3)   log 8

Which implies that

x^(1/3)  = 8     cube both sides

x  = 512

WolframAlpha finds no real solutions to this......

(2^x - 4)^3 + (4^x - 2)^3 = (4^x + 2^x - 6)^3

Let  a  = 2^x    let  a^2  =  (2^x)^2  = (2^2)^x  =  4^x

So we have

( a - 4)^3  + ( a^2 - 2)^3  =  ( a^2 + a - 6)^3

(a - 4)^3  + (a^2 - 2)^3  =  (a^2 + a - 6)^3      write the first two terms as a sum of cubes

[ (a - 4) +( a^2 - 2)]  [  (a - 4)^2 - (a - 4)(a^2 - 2) + (a^2 - 2)^2]  =  (a^2 + a - 6)^3

[ a^2 + a - 6 ]   [ (a - 4)^2 - ( a- 4) (a^2 - 2) + (a^2 - 2)^2 ]  =  [ a^2 + a - 6 ]^3

 a^6 - 6 a^4 + a^3 + 48 a - 72   = a^6 + 3 a^5 - 15 a^4 - 35 a^3 + 90 a^2 + 108 a - 216

3a^5  - 9a^4 - 36a^3 + 90a^2 + 60a - 144  = 0  this factors as

3 (a - 4) (a - 2) (a + 3) (a^2 - 2) = 0

Since a = 2^x....then only the positive roots need be considered

So

a = 4        a   = 2        and    a  = √2

So

2^x  =  4    ⇒  x  = 2

2^x  =  2  ⇒  x = 1

2^x  = √2   ⇒  x = 1/2 

Answer:

x=512

Explanation:

Using the laws of logarithms

Conversion: 3 3/8 = 3 · 8 + 38 = 278

Subtract: 12 - 278 = 1 · 42 · 4 - 278 = 48 - 278 = 4 - 278 = -238
The common denominator you can calculate as the least common multiple of the both denominators - LCM(2, 8) = 8

Multiple: 13 * (-238) = 1 · (-23)3 · 8 = -2324
Multiply both numerators and denominators. Result fraction keep to lowest possible denominator GCD(-23, 24) = 1

I belive I could be wrong.

See if this helps  :

https://web2.0calc.com/questions/please-check-my-geometry-proof-they-hate-me

x^2  =  2012^2(a - b)^2 

y^2 =  2012^2 (b - c)^2

z^2 =  2012^2 (c - a)^2

xy = 2012^2(a - b) (b - c)

yz  = 2012^2 (b -c) (c - a)

xz =  2012^2 (a - b)(c - a)

So

 {x^2 + y^2 + z^2}/{xy + yz + zx}  =

[  2012^2 [  ( a- b)^2  + (b - c)^2 + ( c - a)^2 ] ]/ [ 2012^2 [ (a-b)(b -c) +(b -c)(c - a) +(a -b) (c-a)] ]

[ ( a - b)^2 + (b - c)^2  + (c - a)^2) ]  /  [  (b - c) ( a - b + c - a)  + (a - b)(c - a) ]

[ ( a - b)^2 + (b  - c)^2 + (c - a)^2 ]  /  [  (b -c) (c - b) + ( a - b)(c - a) ] 

[  a^2 - 2ab + b^2 + b^2 - 2bc + c^2 + c^2 - 2ac + a^2]  / [  - ( b - c)^2  + (a - b)(c - a)]

[  2 [ a^2 + b^2 + c^2]  - 2 [ ab + bc + ac ] ]/ [ -b^2 + 2bc - c^2   + ac - bc - a^2 + ab ]

[ 2 [ a^2 + b^2 + c^2] - 2 [ ab + bc + ac] ] / [ - [ a^2 + b^2 + c^2]  + [ ab + bc + ac] ]

[ 2  [ a^2 + b^2 + c^2 - ab - bc - ac ] ]  /  [ - 1 [ a^2 + b^2 + c^2 - ab + bc + ac ] ]

=

2 / -1  =

-2

Solve for x over the real numbers:
(2^x - 4)^3 + (4^x - 2)^3 = (-6 + 2^x + 4^x)^3

After a few expansions and re-arrangements, we have:

The left hand side factors into a product with five terms:
-3 (2^x - 4) (2^x - 2) (2^x + 3) (2^(2 x) - 2) = 0

Divide both sides by -3:
(2^x - 4) (2^x - 2) (2^x + 3) (2^(2 x) - 2) = 0

Split into four equations:
2^x - 4 = 0 or 2^x - 2 = 0 or 2^x + 3 = 0 or 2^(2 x) - 2 = 0

Add 4 to both sides:
2^x = 4 or 2^x - 2 = 0 or 2^x + 3 = 0 or 2^(2 x) - 2 = 0

4 = 2^2:
2^x = 2^2 or 2^x - 2 = 0 or 2^x + 3 = 0 or 2^(2 x) - 2 = 0

Equate exponents of 2 on both sides:
x = 2 or 2^x - 2 = 0 or 2^x + 3 = 0 or 2^(2 x) - 2 = 0

Add 2 to both sides:
x = 2 or 2^x = 2 or 2^x + 3 = 0 or 2^(2 x) - 2 = 0

2 = 2^1:
x = 2 or 2^x = 2^1 or 2^x + 3 = 0 or 2^(2 x) - 2 = 0

Equate exponents of 2 on both sides:
x = 2 or x = 1 or 2^x + 3 = 0 or 2^(2 x) - 2 = 0

Subtract 3 from both sides:
x = 2 or x = 1 or 2^x = -3 or 2^(2 x) - 2 = 0

2^x = -3 has no solution since for all z element R, 2^z>0 and -3<0:
x = 2 or x = 1 or 2^(2 x) - 2 = 0

Add 2 to both sides:
x = 2 or x = 1 or 2^(2 x) = 2

Take the logarithm base 2 of both sides:
x = 2 or x = 1 or 2 x = 1

Divide both sides by 2:
x = 2            or            x = 1         or              x = 1/2

Parallelogram EFGH is similar toparallelogram JKLM.

What is the scale factor of a dilation from  EFGH to JKLM ?

Enter your answer, as a decimal, in the box.

If this is

sqrt(x) + sqrt(y) + sqrt(z) = 10

x + y + z = 38

sqrt (xy ) + sqrt (xz ) + sqrt (yz) = 30

No solutions exist

I  let        a = sqrt (x), b  = sqrt(y)  and c = sqrt (z)

And...as the guest points  out ....   there is a solution  if the last equation = 31

a = 2   b  = 3  and  c = 5

So     x  = 4, y = 9  and z  = 25

I can give you the details of my method...if you want them

Your last equation should be: sqrtxy + sqrtxz + sqrtyz = 31 NOT 30 !!

By simple iteration x=4, y=9, z=25

( z - 3)^4 + (z - 5)^4  = - 8

Let    a + 1  =  z - 3      and  a - 1  = z  - 5   

So we have

(a + 1)^4  +  ( a - 1)^4  = - 8

a^4 + 4a^3 + 6a^2 + 4a + 1  + a^4 - 4a^3 + 6a^2 - 4a + 1  =  - 8

2a^4 +  12a^2  + 2  =  - 8

2a^4  + 12a^2  + 10  =  0

a^4 + 6a^2  + 5  =  0        factor

(a^2 + 5) ( a^2 + 1)  = 0

Setting each factor to 0  and solving for a we have that

a = i√5 , -i√5, i , -i

So...

a =  z - 4  ⇒  a + 4  = z

z =   4 + i√5,  4 - i√5,  4 + i,   4 - i

Yep.....yours makes better sense.....!!!

I think it is better that you state what the problem, you are trying to solve, is and then we might be able to assist you. The top formula: PV - [FV/(1+i)^N] will always give you a zero, IF you are working on the same problem, because this part [FV/(1+i)^N] gives you PV!!. So, in essence, you are subtracting PV -PV =0 !!. If you are trying to write a code to solve for PV, FV, PMT, i, N, I may refer you to this online financial calculator, which not only can solve for any of the above 5 variables, but also provides the mathematical formulas programmed into it. Here it is: 

https://arachnoid.com/finance/