The power of 3!

Let $y = 3^{57x}$ , so the given expression simplifies to $g(y) = y^3 - 243y^2 + 729y - 27 = 0$ . We can check that $g(0) < 0, g(1) > 0, g(3) =0, g(81) < 0, g (\infty)>0$ , hence we have 3 real roots that are greater than 0.

Let $y_1$ , $y_2$ and $y_3$ be the solutions, thus the corresponding solutions to $f(x)$ are $x_1 = \frac{\log_3 y_1}{57}$ , $x_2 = \frac{\log_3 y_2}{57}$ and $x_3 = \frac{\log_3 y_3}{57}$ . By Vieta's formulae, we have $y_1y_2y_3 = 27$ . Taking $\log_3$ of both sides, we have $\begin{aligned} \log_3 y_1 + \log_3 y_2 + \log_3 y_3 &= 3 \\ 57(x_1 + x_2 + x_3) &= 3\\ x_1 + x_2 + x_3 &= \frac{1}{19}\\ \end{aligned}$

Thus $S = \frac{1}{19}$ , hence $\frac{1}{S} = 19$ .

For roots of the equation, set f(x) = 0

$f(x) = 3^{171x-3} - 3^{114x+2} + 3^{57x+3} - 1 = 0 \\ 3^{-3}(3^{57x})^3 - 3^2(3^{57x})^2 + 3^3(3^{57x}) - 1 = 0 \\ (3^{57x})^3 - 3^5(3^{57x})^2 + 3^6(3^{57x}) - 3^3 = 0 \\ (3^{57x} - 3)((3^{57x})^2 + 3(3^{57x}) + 9) - 3^5(3^{57x})(3^{57x} - 3) = 0 \\ (3^{57x} - 3)(3^{57x})^2 + 3(3^{57x}) + 9 - 3^5(3^{57x}) = 0 \\ (3^{57x} - 3)((3^{57x})^2 - 240(3^{57x}) + 9) = 0$

From the first factor, $3^{57x} = 3^1 \Rightarrow x = 1/57$ .

For the second factor, it is a quadratic with the real roots on $3^{57x}$ because the discriminant is $240^2 - 4*1*9 > 0$ . The product of these roots {a, b} is 9 by Vieta's formula.

$3^{57a}3^{57b}=3^2 \\ 3^{57(a+b)}=3^2 \\ a + b = 2/57 \\ \\ \therefore S = x + a + b = 1/57 + 2/57 = 1/19 \Rightarrow \boxed{1/S = 19}$

Note: I didn't calculate all three roots directly. This was intentional as it was not necessary (and it saved some time too).

Let f(x) = 3^(171x−3) − 3^(114x+2) + 3^(57x+3) − 1

Since S is the sum of all real values of x that satisfy f(x) = 0, all we do is let f(x) = 0.

We now have 0 = 3^(171x−3) − 3^(114x+2) + 3^(57x+3) − 1.

By the properties of exponents, 0 = (3^(57x)(3))/27 - (3^(57x)(2))(9) + (3^(57x))(27) - 1

Let y = 3^57x

The equation then becomes 0 = (1/27)y^3 - 9y^2 + 27y - 1

Multiplying everything by 27 to remove the denominator 0 = y^3 - 243y^2 + 729y - 27

Factoring the polynomial we get 0 = (y-3)(y^2-240y+9)

Solving for y, we get y = {3, 120 + sqrt(14391), 120 - sqrt(14391)}

Solving for x by reverse substitution, 3^57x = 3 implies that x = 1/57; 3^57x = 120 + sqrt(14391) implies that x = (1/57) log (120 + sqrt(14391)) to the base 3; and 3^57x = 120 - sqrt(14391) implies that x = (1/57) log (120 - sqrt(14391)) to the base 3.

Adding the 2nd and 3rd values of x and applying the laws of logarithms, we get (1/57) log 9 to the base 3 or 2/57 since

(120 + sqrt(14391) (120 - sqrt(14391) = 14400 - 14391 = 9

Then adding the first value of x to the sum of the latter 2 we get 1/19

Finally, solving for 1/S we simply get the reciprocal of the sum which is 19.

Given that f(x) = 3^(171x-3)-3^(114x+2)+3^(57x+3)-1 and S be the sum of all real values of x that satisfy that f(x)= 0 First , equate f(x) =0 , then 3^(171x-3)-3^(114x+2)+3^(57x+3)-1=0 divide the first term to 3^(-6) and the third term to 3^(-2) , then the resulting equation will be (3^(-6))(3^(171x+3))-3^(114x+2)+(3^(-2))(3^(57x+1))-1=0

Let y=3^(57x+1) by substitution (3^(-6))(y^3)-(y^2)-(3^(-2))y-1=0 divide both sides by 3^(-6) y^3-729y^2-81y-729=0

The product of the roots of a cubic polynomial is equal to (-d/a). Then, the product of the roots of y^3-729y^2-81y-729=0 is 729

since the function has three root , then Let a,b,c be the roots of the function.

(3^(57a+1))(3^(57b+1))(3^(57c+1))=729 (3^((57)(a+b+c)+3))=3^6 since a+b+c = S and apply the Laws of Exponents 57S+3 = 6 57S= 3 Multiplying both sides by 3/S, therefore 1/S=19

(a^3)-243a+729a-27=0 or (a-3)(a^2-240a+9)=0 gives a=3, a=120+sqrt(14391) and a=120-sqrt(14391) Then, x=1/57, x=(1/57)log{120+sqrt(14391)}, x=(1/57)log{120-sqrt(14391)} sum of the values of x, S=(1/57)+(1/57)log{120+sqrt(14391)}+(1/57)log{120-sqrt(14391)} =(1/57)+(1/57)log9 [Taking the logarithm with base 3 in all cases]. =(1/57)+2(1/57)=(1/19). Therefore (1/S)=19.

Rewrite f(x) as f(x) = \frac{1}{27} 3^{171x} - 9 *3^{114x} + 27 * 3^{57x} - 1.

Now let y = 3^{57x}.

Now f(x) = \frac{1}{27}y^3 - 9y^2 + 27y - 1.

Suppose the roots to this cubic are r, s, and t.

Thus r = 3^{57x}, s = 3^{57x}, and t = 3^{57x} for some values of x.

Taking logs of both sides, we get log 3 r = 57x, log 3 s = 57x, and log_3 t = 57x.

Dividing by 57, we get that \frac{log 3r}{57} = x, \frac{log 3s}{57} = x, and \frac{log 3t}{57} = x. We want the sum of all possible values of x, so we add all the logs together, getting \frac{log 3(rst)}{57} = S. rst = -\frac{-1}{\frac{1}{27}} from the cubic equation, thus the equation becomes

\frac{log_3 27}{57} = S, \frac{3}{57} = S \frac{57}{3} = \frac{1}{S}

19 = \frac{1}{S}.

Let $y=3^{57x}, x = \frac {\log_3 y}{57}$ . We have $f(y)=\frac{y^3}{27} -9y^2 +27y-1$ which implies that $y^3-3^5y^2+3^6y-3^3=0$ . The product of solutions for $y$ is $3^3$ by Vieta formula. So, we have $S = x_1 + x_2 + x_3 = \sum_y{ \frac{\log_3{y}}{57} }= \frac{\log_3{y_1y_2y_3}}{57}= \frac{3}{57}$ (due to the property of logarithms). So $\frac{1}{S}= \frac{57}{3}=19$ .

[Edits for clarity - Calvin]

Consider that 3^{57x} as p, then:

[\frac {1}{27}p^{3}-9p^{2}+27p-1=0]× 27

27p^{3}-243p^{2}+729p-27=0

From the theory of poly nominal, it is known a formula for a poly nominal with degrees of 3:

p {1}×p {2}×p_{3}= /frac {-d}{a}

Then 3^{57(x {1}+x {2}+x_{3})}=27

Since x {1}+x {2}+x_{3}= S, then:

3^{57S}=27

Put the equation of log for both side so that:

57S log3= 3log3

Then S= /frac {3}{57} and /frac {1}{S}=19

1st: 3^(171x-3)=1 3^(171x-3)=3^0 171x-3=0 171x=3 x=1/57 2nd: 3^(114x+2)=1 3^(114x+2)=3^0 114x+2=0 x=-1/57 3rd: 3^(57x+3)=1 3^(57x+3)=3^0 57x+3=0 x=1/19 then: 0=3^0-3^0+3^0-1 0=0 therefore: 1/s=1(1/57-1/57+1/19) 1/s=19