Diketahui f(x)=2^(5-x)+2x=^x-12. Jika f(x1)=f(x2)=0 maka

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Pembahasan

Untuk menyelesaikan soal ini, kita perlu mencari nilai x1 dan x2 yang memenuhi persamaan f(x) = 2^(5-x) + 2^(x-1) - 12 = 0, lalu menghitung hasil perkaliannya (x1*x2).

Misalkan y = 2^x.
Maka 2^(5-x) dapat ditulis sebagai 2^5 * 2^(-x) = 32 * (1/2^x) = 32/y.
Dan 2^(x-1) dapat ditulis sebagai 2^x * 2^(-1) = y * (1/2) = y/2.

Substitusikan ke dalam persamaan f(x):
32/y + y/2 - 12 = 0

Untuk menghilangkan penyebut, kalikan seluruh persamaan dengan 2y:
2y * (32/y) + 2y * (y/2) - 2y * 12 = 0
64 + y^2 - 24y = 0

Susun ulang persamaan menjadi bentuk kuadrat standar:
y^2 - 24y + 64 = 0

Kita dapat menggunakan rumus Vieta untuk persamaan kuadrat ay^2 + by + c = 0, di mana akar-akarnya adalah y1 dan y2. Hasil perkalian akar-akarnya adalah y1 * y2 = c/a.
Dalam kasus ini, a=1, b=-24, dan c=64. Maka, y1 * y2 = 64/1 = 64.

Ingat bahwa kita memisalkan y = 2^x. Jadi, jika f(x1) = 0 dan f(x2) = 0, maka y1 = 2^x1 dan y2 = 2^x2.

Kita memiliki y1 * y2 = 64.
Substitusikan kembali y1 dan y2:
2^x1 * 2^x2 = 64

Menggunakan sifat eksponen (a^m * a^n = a^(m+n)):
2^(x1 + x2) = 64

Karena 64 = 2^6, maka:
2^(x1 + x2) = 2^6

Ini berarti x1 + x2 = 6.

Namun, soal menanyakan hasil perkalian x1*x2, bukan jumlahnya (x1+x2). Mari kita cek kembali apakah ada informasi yang terlewat atau jika ada kesalahan dalam asumsi.

Persamaan kuadrat dalam y adalah y^2 - 24y + 64 = 0. Kita perlu mencari akar-akarnya terlebih dahulu untuk mendapatkan x1 dan x2.
Kita bisa menggunakan rumus kuadratik y = [-b ± sqrt(b^2 - 4ac)] / 2a.

y = [24 ± sqrt((-24)^2 - 4 * 1 * 64)] / (2 * 1)
y = [24 ± sqrt(576 - 256)] / 2
y = [24 ± sqrt(320)] / 2
y = [24 ± sqrt(64 * 5)] / 2
y = [24 ± 8 * sqrt(5)] / 2
y = 12 ± 4 * sqrt(5)

Jadi, y1 = 12 + 4 * sqrt(5) dan y2 = 12 - 4 * sqrt(5).

Sekarang, kita punya:
2^x1 = 12 + 4 * sqrt(5)
2^x2 = 12 - 4 * sqrt(5)

Untuk mencari x1*x2, kita bisa menggunakan logaritma:
x1 = log2(12 + 4 * sqrt(5))
x2 = log2(12 - 4 * sqrt(5))

x1 * x2 = log2(12 + 4 * sqrt(5)) * log2(12 - 4 * sqrt(5))

Ini terlihat rumit. Mari kita periksa kembali apakah ada cara yang lebih sederhana atau jika ada properti yang bisa digunakan.

Kita tahu bahwa y1 * y2 = 64.
Dan 2^x1 * 2^x2 = 64
2^(x1 + x2) = 64
x1 + x2 = 6.

Soal yang diberikan adalah f(x)=2^(5-x)+2^(x-1)-12. Jika f(x1)=f(x2)=0 maka x1x2=....

Perhatikan kembali persamaan y^2 - 24y + 64 = 0, di mana y = 2^x.
Ini berarti kita mencari nilai x sedemikian rupa sehingga 2^x adalah akar dari persamaan kuadrat tersebut.

Jika kita memiliki sebuah fungsi g(x) = a^x, dan kita memiliki persamaan kuadrat dalam bentuk g(x), misalnya P(g(x)) = 0, dan akar-akar persamaan kuadrat tersebut adalah g(x1) dan g(x2), maka:

Dalam kasus ini, kita memiliki persamaan dalam 2^x, yaitu (2^x)^2 - 24(2^x) + 64 = 0. (Ini salah, karena 2^(5-x) bukan (2^x)^2).

Mari kita kembali ke:
32/y + y/2 - 12 = 0
y^2 - 24y + 64 = 0, dengan y = 2^x.

Jika y1 dan y2 adalah akar-akar dari persamaan kuadrat ini, maka:
y1 = 2^x1 dan y2 = 2^x2.

Kita tahu dari sifat logaritma bahwa jika a = b^c, maka c = log_b(a).
Jadi, x1 = log_2(y1) dan x2 = log_2(y2).

Maka, x1 * x2 = log_2(y1) * log_2(y2).

Kita juga tahu bahwa y1 * y2 = 64.

Mari kita lihat sifat logaritma:
log(a*b) = log(a) + log(b)
log(a/b) = log(a) - log(b)
log(a^n) = n*log(a)

Kita memiliki x1 + x2 = log_2(y1) + log_2(y2) = log_2(y1 * y2) = log_2(64) = 6.
Ini konsisten.

Untuk mencari x1*x2, kita perlu nilai dari log_2(y1) dan log_2(y2).

Mari kita gunakan kembali akar-akar yang sudah kita cari:
y1 = 12 + 4 * sqrt(5)
y2 = 12 - 4 * sqrt(5)

x1 = log_2(12 + 4 * sqrt(5))
x2 = log2(12 - 4 * sqrt(5))

x1 * x2 = log2(12 + 4 * sqrt(5)) * log2(12 - 4 * sqrt(5))

Apakah ada cara lain untuk menyederhanakan ini?

Consider the original equation f(x) = 2^(5-x) + 2^(x-1) - 12 = 0.
Let x1 and x2 be the roots. Then f(x1) = 0 and f(x2) = 0.

Let's consider a property of exponential equations. If $a^{f(x)} + a^{g(x)} = c$ and there are two roots $x_1, x_2$ such that $f(x_1) = g(x_2)$ and $f(x_2) = g(x_1)$, then $x_1+x_2$ is constant. This is not directly applicable here.

Let's re-examine the substitution. Let $u = 2^x$. Then $2^{5-x} = 2^5 / 2^x = 32/u$. And $2^{x-1} = 2^x / 2^1 = u/2$. The equation becomes $32/u + u/2 - 12 = 0$. Multiplying by $2u$ gives $64 + u^2 - 24u = 0$, or $u^2 - 24u + 64 = 0$.
The roots of this quadratic equation are $u_1 = 2^{x_1}$ and $u_2 = 2^{x_2}$.
From Vieta's formulas for the quadratic equation $au^2 + bu + c = 0$, the product of the roots is $u_1 u_2 = c/a$. In our case, $a=1, b=-24, c=64$, so $u_1 u_2 = 64/1 = 64$.
Substituting back $u_1 = 2^{x_1}$ and $u_2 = 2^{x_2}$, we get $2^{x_1} 	imes 2^{x_2} = 64$. Using the property $a^m 	imes a^n = a^{m+n}$, we have $2^{x_1 + x_2} = 64$. Since $64 = 2^6$, we get $x_1 + x_2 = 6$.

The question asks for $x_1 x_2$, not $x_1 + x_2$. This indicates that either the question is ill-posed, there is a typo, or there's a property I'm missing.

Let's consider the structure of the original function: $f(x) = 2^{5-x} + 2^{x-1} - 12$. 
Let's test symmetry. If we replace $x$ with $6-x$, what do we get?
$f(6-x) = 2^{5-(6-x)} + 2^{(6-x)-1} - 12$
$f(6-x) = 2^{5-6+x} + 2^{5-x} - 12$
$f(6-x) = 2^{x-1} + 2^{5-x} - 12$

We see that $f(6-x) = f(x)$.
This means that if $x_1$ is a root, then $6-x_1$ is also a root.
So, the two roots $x_1$ and $x_2$ must satisfy $x_2 = 6-x_1$ (or $x_1 = 6-x_2$).

Therefore, $x_1 + x_2 = x_1 + (6-x_1) = 6$. This confirms our earlier finding about the sum of the roots.

Now, we need to find $x_1 x_2$. Since $x_2 = 6-x_1$, we have $x_1 x_2 = x_1 (6-x_1) = 6x_1 - x_1^2$.
This expression still depends on $x_1$. We need a numerical value.

Let's reconsider the quadratic equation for $u=2^x$: $u^2 - 24u + 64 = 0$. The roots are $u_1 = 2^{x_1}$ and $u_2 = 2^{x_2}$.
We found $u_1 = 12 + 4	ext{sqrt}(5)$ and $u_2 = 12 - 4	ext{sqrt}(5)$.

$x_1 = 	ext{log}_2(12 + 4	ext{sqrt}(5))$
$x_2 = 	ext{log}_2(12 - 4	ext{sqrt}(5))$

$x_1 x_2 = 	ext{log}_2(12 + 4	ext{sqrt}(5)) 	imes 	ext{log}_2(12 - 4	ext{sqrt}(5))$.

Could there be a mistake in the problem statement or the expected answer format? The problem asks for a specific value for $x_1 x_2$.

Let's verify the symmetry property again. If $f(x) = a^{k-x} + a^{x-m} - c$, and $k-x_1 = x_2-m$ and $k-x_2 = x_1-m$, then $x_1+x_2 = k+m$. Here, $f(x) = 2^{5-x} + 2^{x-1} - 12$. So $k=5, m=1$. The symmetry check was $f(6-x) = 2^{5-(6-x)} + 2^{(6-x)-1} - 12 = 2^{x-1} + 2^{5-x} - 12 = f(x)$. So, if $x_1$ is a root, $6-x_1$ is also a root. This implies $x_1+x_2=6$. This is correct.

Is there a transformation that turns $x_1 x_2$ into a known constant from the quadratic equation of $u=2^x$? Not directly.

Let's consider the possibility that the question implicitly assumes specific properties or that there's a known identity related to this form.

If the question was about $x_1+x_2$, the answer would be 6.

Let's check if the roots $u_1, u_2$ have a specific relationship that simplifies the product of their logarithms. For example, if $u_1 = u_2^k$ or something similar.

$u_1 = 12 + 4	ext{sqrt}(5)$
$u_2 = 12 - 4	ext{sqrt}(5)$

$u_1 	imes u_2 = (12 + 4	ext{sqrt}(5))(12 - 4	ext{sqrt}(5)) = 12^2 - (4	ext{sqrt}(5))^2 = 144 - (16 	imes 5) = 144 - 80 = 64$.
This is consistent. $u_1 u_2 = 64$.

Let's assume the question meant to ask for something that can be derived from $u_1 u_2$. The product $x_1 x_2$ is not directly related to $u_1 u_2$ in a simple way without knowing the individual values of $u_1$ and $u_2$.

Could the question be about $x_1$ and $x_2$ in a different base? No, the base is clearly 2.

Let's review the problem and common patterns in such questions. Often, when $f(x) = f(c-x)$ for some constant $c$, it implies the roots are symmetric around $c/2$. Here, $c=6$, so the midpoint is 3. If $x_1=3$, then $x_2=3$. In that case $x_1 x_2 = 9$. Let's check if $x=3$ is a root.
$f(3) = 2^{5-3} + 2^{3-1} - 12 = 2^2 + 2^2 - 12 = 4 + 4 - 12 = 8 - 12 = -4 
eq 0$.
So, $x=3$ is not a root, and $x_1 
eq x_2$.

If the question intends a numerical answer, there might be a simpler structure or a misunderstanding of the question. 

Let's consider the possibility that the problem setter made a mistake and intended for $x_1 x_2$ to be derivable from the coefficients of the quadratic in $u$. This is unlikely as the product of roots of the exponential function is directly related to the sum of the exponents.

Let's assume there might be a property related to the product of logarithms of roots of a quadratic.

If we express $u_1$ and $u_2$ in a different form: $u = 12 \pm 4\sqrt{5}$. This doesn't simplify the log. 

Could the question be interpreted differently? "Jika f(x1)=f(x2)=0 maka x1x2=...."
This implies that there are exactly two distinct roots $x_1$ and $x_2$.

Let's consider the function $g(u) = u^2 - 24u + 64$. The roots are $u_1, u_2$. Let $h(x) = 2^x$. Then the roots of $g(h(x)) = 0$ are $x_1, x_2$ such that $h(x_1) = u_1$ and $h(x_2) = u_2$. So $2^{x_1} = u_1$ and $2^{x_2} = u_2$.

We know $u_1 u_2 = 64$. Thus $2^{x_1} 2^{x_2} = 2^{x_1+x_2} = 64$, which gives $x_1+x_2 = 6$. 

If the question intends a specific numerical value for $x_1 x_2$, and given the symmetry $f(x) = f(6-x)$, it's possible that the problem is constructed such that $x_1 x_2$ has a specific value related to the symmetry constant.

Consider a function like $f(x) = (x-a)(x-b)$. Then $x_1=a, x_2=b$ and $x_1x_2 = ab$. 

In our case, we have $2^{x_1+x_2} = 64$. This is about the sum. 

Let's consider the original problem source or context if available. Without it, we must rely on the given information.

If this were a multiple-choice question, the options might provide a clue. 

Let's re-read the original equation carefully: $f(x)=2^(5-x)+2^(x-1)-12=0$. 
This is $2^5 	imes 2^{-x} + 2^x 	imes 2^{-1} - 12 = 0$. 
$32/2^x + 2^x/2 - 12 = 0$. 
Let $y = 2^x$. Then $32/y + y/2 - 12 = 0$. 
Multiply by $2y$: $64 + y^2 - 24y = 0$. 
$y^2 - 24y + 64 = 0$. 
Roots are $y_1 = 2^{x_1}$ and $y_2 = 2^{x_2}$. 
$y_1 y_2 = 64$. 
$2^{x_1} 2^{x_2} = 2^{x_1+x_2} = 64$. 
$x_1+x_2 = 6$. 

Is there any special case where $x_1 x_2$ can be found directly?

Let's consider the derivative to find the extremum, which might give insight into the roots. However, we are given that there are two roots $x_1, x_2$. 

Could the question be flawed? It's a possibility. If the question meant to ask for $x_1+x_2$, the answer is 6.

Let's search for similar problems online or in textbooks. Problems involving $a^{f(x)} + a^{g(x)} = c$ often lead to a quadratic in $a^x$. The relation between roots of the quadratic and the exponents is usually about the sum of exponents.

If $f(x) = k rac{a^{mx}}{b^{nx}} + l rac{b^{px}}{a^{qx}} = c$, then sometimes a transformation can lead to a symmetric equation where if $x_0$ is a root, then $m-x_0$ is also a root.

In our case, the structure is $2^{5-x}$ and $2^{x-1}$. The exponents are $5-x$ and $x-1$. Their sum is $(5-x) + (x-1) = 4$. This does not seem directly related to the sum of roots $x_1+x_2=6$. 

However, the symmetry $f(x) = f(6-x)$ is strong. It means the roots are symmetric around $x=3$. 

Let's consider the possibility of a typo in the original question. For example, if the function was $2^{x} + 2^{6-x} = c$, then if $x_1$ is a root, $6-x_1$ is also a root, and $x_1+x_2 = 6$. The product $x_1 x_2$ would still be dependent on the specific value of $x_1$. 

For instance, if $2^x + 2^{6-x} = 20$, let $y=2^x$. $y + 64/y = 20$. $y^2 - 20y + 64 = 0$. $(y-4)(y-16) = 0$. So $y=4$ or $y=16$. 
$2^x = 4 
ightarrow x=2$. 
$2^x = 16 
ightarrow x=4$. 
Here, $x_1=2, x_2=4$. $x_1+x_2 = 6$. $x_1 x_2 = 8$. 

In our problem, $y^2 - 24y + 64 = 0$. The roots are $y_1 = 12 + 4	ext{sqrt}(5)$ and $y_2 = 12 - 4	ext{sqrt}(5)$. 
$2^{x_1} = 12 + 4	ext{sqrt}(5)$
$2^{x_2} = 12 - 4	ext{sqrt}(5)$

Numerically, $	ext{sqrt}(5) \approx 2.236$.
$y_1 \approx 12 + 4(2.236) = 12 + 8.944 = 20.944$.
$y_2 \approx 12 - 4(2.236) = 12 - 8.944 = 3.056$.

$x_1 = 	ext{log}_2(20.944) \approx 4.388$
$x_2 = 	ext{log}_2(3.056) \approx 1.611$

$x_1 + x_2 
esse 4.388 + 1.611 = 5.999 
esse 6$. (Matches our sum).

$x_1 x_2 
esse 4.388 	imes 1.611 \approx 7.066$.
This is not a simple integer or common fraction.

Given that this is likely a contest math problem or textbook problem, there is usually a clean answer. The fact that $x_1+x_2=6$ is derived cleanly from the coefficients of the quadratic in $y=2^x$ suggests this is intended. The product $x_1 x_2$ is not directly obtainable. 

Could the question be about finding the product of the roots of the quadratic in $y$? No, it specifically asks for $x_1x_2$. 

Let's reconsider the symmetry $f(x) = f(6-x)$. This implies that the roots $x_1$ and $x_2$ are located such that their midpoint is 3. If $x_1 
eq x_2$, then $x_1 = 3 - d$ and $x_2 = 3 + d$ for some distance $d > 0$. 
Then $x_1 + x_2 = (3-d) + (3+d) = 6$. 
And $x_1 x_2 = (3-d)(3+d) = 9 - d^2$. 

We need to find $d^2$.
$2^{x_1} = 2^{3-d} = 2^3 / 2^d = 8 / 2^d = y_1$
$2^{x_2} = 2^{3+d} = 2^3 	imes 2^d = 8 	imes 2^d = y_2$

$y_1 y_2 = (8/2^d)(8 	imes 2^d) = 64$. This is consistent. 

We have $y_1 = 12 + 4	ext{sqrt}(5)$ and $y_2 = 12 - 4	ext{sqrt}(5)$.
So, $8 / 2^d = 12 - 4	ext{sqrt}(5)$ and $8 	imes 2^d = 12 + 4	ext{sqrt}(5)$.

Let's use the second equation to find $2^d$: $2^d = (12 + 4	ext{sqrt}(5)) / 8 = (3 + 	ext{sqrt}(5)) / 2$.

Now we need $d^2$. From $2^d = (3 + 	ext{sqrt}(5)) / 2$, we can find $d$ using logarithms: $d = 	ext{log}_2((3 + 	ext{sqrt}(5)) / 2)$. This is getting complicated.

Alternatively, we need $d^2$ for $9-d^2$. 
Let's check if $y_1$ or $y_2$ can be expressed in terms of powers of 2 with a base related to $3 	ext{ and } 	ext{sqrt}(5)$.

Consider $(rac{1+	ext{sqrt}(5)}{2})^2 = rac{1 + 2	ext{sqrt}(5) + 5}{4} = rac{6 + 2	ext{sqrt}(5)}{4} = rac{3 + 	ext{sqrt}(5)}{2}$.
This is the golden ratio squared ($\phi^2$).
So, $2^d = rac{3 + 	ext{sqrt}(5)}{2} = 	ext{golden ratio}^2 = rac{1+	ext{sqrt}(5)}{2}^2$.

This means $2^d = 	ext{golden ratio}^2$. 
$d = 	ext{log}_2(	ext{golden ratio}^2) = 2 	ext{log}_2(	ext{golden ratio})$.

Then $d^2 = (2 	ext{log}_2(	ext{golden ratio}))^2 = 4 (	ext{log}_2(	ext{golden ratio}))^2$.

$x_1 x_2 = 9 - d^2 = 9 - 4 (	ext{log}_2(rac{1+	ext{sqrt}(5)}{2}))^2$.
This is still not a simple number.

Is there any mistake in my interpretation of the symmetry or the substitution?

Let's revisit the possibility of a common error or a trick. 
If the base was variable, say $a^{5-x} + a^{x-1} - 12 = 0$, the relationship between roots would depend on $a$. 

Consider the possibility that the question implies $x_1$ and $x_2$ are related in a way that $x_1 x_2$ is a constant derived from the structure of the original equation, not just the quadratic in $y$. 

Let's check if there's a scenario where $x_1 x_2$ is related to the constant terms in the exponents. 
Exponents: $5-x$ and $x-1$. 
Constants: 5 and -1. 
Sum of constants: $5 + (-1) = 4$. 
Product of constants: $5 	imes (-1) = -5$. 

If we consider $x_1$ and $x_2$ such that $5-x_1 = x_2-1$, then $x_1+x_2 = 6$. This is the same symmetry we found.

What if we consider the property of the roots of the equation $u^2 - 24u + 64 = 0$? 
We have $u_1 = 2^{x_1}$ and $u_2 = 2^{x_2}$. 

Perhaps there's a typo in the question and it should lead to a quadratic in $y$ where the roots are powers of 2 that are easy to work with.
For example, if $u^2 - 20u + 64 = 0$, roots are 4 and 16. $2^{x_1}=4 
ightarrow x_1=2$. $2^{x_2}=16 
ightarrow x_2=4$. Then $x_1 x_2 = 8$. 
This would correspond to $f(x) = 2^{5-x} + 2^{x-1} - 20 = 0$. No, that's not it. 
If $2^x + 2^{6-x} = 20$, then $y + 64/y = 20$, $y^2-20y+64=0$. Roots are $y=4, 16$. $2^x=4 
ightarrow x=2$. $2^x=16 
ightarrow x=4$. $x_1=2, x_2=4$. $x_1+x_2=6$. $x_1x_2=8$. 
This form $2^x + 2^{c-x}$ leads to $x_1+x_2 = c$. 

Our original form is $2^{5-x} + 2^{x-1} - 12 = 0$. 
Let's transform the exponents to have a common variable. Let $z = x-1$. Then $x = z+1$. And $5-x = 5-(z+1) = 4-z$.
So, $2^{4-z} + 2^z - 12 = 0$. 
Let $w = 2^z$. Then $2^4/w + w - 12 = 0$. 
$16/w + w - 12 = 0$. 
Multiply by $w$: $16 + w^2 - 12w = 0$. 
$w^2 - 12w + 16 = 0$. 
Let the roots be $w_1 = 2^{z_1}$ and $w_2 = 2^{z_2}$. 
$w_1 w_2 = 16$. 
$2^{z_1} 2^{z_2} = 2^{z_1+z_2} = 16$. 
$z_1+z_2 = 4$. 

Recall $z = x-1$. So $z_1 = x_1-1$ and $z_2 = x_2-1$. 
$(x_1-1) + (x_2-1) = 4$. 
$x_1 + x_2 - 2 = 4$. 
$x_1 + x_2 = 6$. This is consistent. 

Now, we need $x_1 x_2$. 
$z_1 = x_1-1$, so $x_1 = z_1+1$. 
$z_2 = x_2-1$, so $x_2 = z_2+1$. 

$x_1 x_2 = (z_1+1)(z_2+1) = z_1 z_2 + z_1 + z_2 + 1$. 

We know $z_1 + z_2 = 4$. We need $z_1 z_2$. 
From the quadratic equation $w^2 - 12w + 16 = 0$, the product of roots is $z_1 z_2 = 16/1 = 16$. 

So, $x_1 x_2 = z_1 z_2 + (z_1 + z_2) + 1 = 16 + 4 + 1 = 21$.

Let's verify this. The roots of $w^2 - 12w + 16 = 0$ are $w = rac{12 	ext{ ± } 	ext{sqrt}(144 - 4 	imes 16)}{2} = rac{12 	ext{ ± } 	ext{sqrt}(144 - 64)}{2} = rac{12 	ext{ ± } 	ext{sqrt}(80)}{2} = rac{12 	ext{ ± } 4	ext{sqrt}(5)}{2} = 6 	ext{ ± } 2	ext{sqrt}(5)$.

So $w_1 = 6 + 2	ext{sqrt}(5)$ and $w_2 = 6 - 2	ext{sqrt}(5)$.

$2^{z_1} = 6 + 2	ext{sqrt}(5)$
$2^{z_2} = 6 - 2	ext{sqrt}(5)$

We need to find $z_1$ and $z_2$ such that $2^{z_1+z_2} = (6 + 2	ext{sqrt}(5))(6 - 2	ext{sqrt}(5)) = 36 - (4 	imes 5) = 36 - 20 = 16$. This matches $z_1+z_2=4$.

Now, $x_1 = z_1+1$ and $x_2 = z_2+1$. 
$x_1 x_2 = (z_1+1)(z_2+1) = z_1 z_2 + z_1 + z_2 + 1$. 
We found $z_1 z_2 = 16$ and $z_1+z_2 = 4$. 
So, $x_1 x_2 = 16 + 4 + 1 = 21$.

This looks like a correct derivation. The key was the transformation $z=x-1$ to make the exponents symmetric around 0 ($z$ and $-z$ if the base was the same, but here it's $z$ and $4-z$). The transformation $z=x-1$ leads to $2^{4-z} + 2^z - 12 = 0$. Let $w = 2^z$. Then $2^4/w + w - 12 = 0$, leading to $w^2 - 12w + 16 = 0$. The roots are $w_1=2^{z_1}$ and $w_2=2^{z_2}$. We need $x_1 x_2 = (z_1+1)(z_2+1) = z_1z_2 + (z_1+z_2) + 1$. From the quadratic, $z_1z_2 = 16$ and $z_1+z_2 = 12$. Wait, sum of roots of $w^2 - 12w + 16 = 0$ is $w_1+w_2 = 12$. Not $z_1+z_2$. The sum of exponents $z_1+z_2$ is 4, as derived from $2^{z_1+z_2}=16$. This is consistent. The product of exponents $z_1 z_2$ is what we need. 

Let's recheck. We have $w_1 = 2^{z_1}$ and $w_2 = 2^{z_2}$. So $z_1 = 	ext{log}_2(w_1)$ and $z_2 = 	ext{log}_2(w_2)$. 
$z_1 z_2 = 	ext{log}_2(w_1) 	ext{log}_2(w_2)$. This is not $w_1 w_2$. 

The earlier conclusion that $x_1 x_2 = 21$ was based on assuming $z_1 z_2$ was the constant term of the quadratic for $z$. This is incorrect. $z_1$ and $z_2$ are the exponents, not the base variable. 

Back to $y^2 - 24y + 64 = 0$ where $y=2^x$. Roots $y_1=2^{x_1}$ and $y_2=2^{x_2}$. 
$y_1 y_2 = 64 
ightarrow x_1+x_2 = 6$. 
$y_1 = 12 + 4	ext{sqrt}(5)$ and $y_2 = 12 - 4	ext{sqrt}(5)$. 

$x_1 = 	ext{log}_2(12 + 4	ext{sqrt}(5))$
$x_2 = 	ext{log}_2(12 - 4	ext{sqrt}(5))$

$x_1 x_2 = 	ext{log}_2(12 + 4	ext{sqrt}(5)) 	imes 	ext{log}_2(12 - 4	ext{sqrt}(5))$.

Is it possible that $12 	ext{ ± } 4	ext{sqrt}(5)$ can be expressed as powers of 2 in a way that simplifies the log product?
$12 + 4	ext{sqrt}(5) = 2(6 + 2	ext{sqrt}(5))$. 
$12 - 4	ext{sqrt}(5) = 2(6 - 2	ext{sqrt}(5))$. 

So, $x_1 = 	ext{log}_2(2(6 + 2	ext{sqrt}(5))) = 1 + 	ext{log}_2(6 + 2	ext{sqrt}(5))$.
$x_2 = 	ext{log}_2(2(6 - 2	ext{sqrt}(5))) = 1 + 	ext{log}_2(6 - 2	ext{sqrt}(5))$.

$x_1 x_2 = (1 + 	ext{log}_2(6 + 2	ext{sqrt}(5))) (1 + 	ext{log}_2(6 - 2	ext{sqrt}(5)))$.
$x_1 x_2 = 1 + 	ext{log}_2(6 + 2	ext{sqrt}(5)) + 	ext{log}_2(6 - 2	ext{sqrt}(5)) + 	ext{log}_2(6 + 2	ext{sqrt}(5)) 	ext{log}_2(6 - 2	ext{sqrt}(5))$.

Using $a+b = 6$ and $ab = 16$ for logs (from $w^2-12w+16=0$ where roots were $6 	ext{ ± } 2	ext{sqrt}(5)$).
Let $L_1 = 	ext{log}_2(6 + 2	ext{sqrt}(5))$ and $L_2 = 	ext{log}_2(6 - 2	ext{sqrt}(5))$.
$L_1 + L_2 = 	ext{log}_2((6 + 2	ext{sqrt}(5))(6 - 2	ext{sqrt}(5))) = 	ext{log}_2(16) = 4$.

So, $x_1 x_2 = 1 + (L_1 + L_2) + L_1 L_2 = 1 + 4 + L_1 L_2 = 5 + L_1 L_2$.

We need $L_1 L_2 = 	ext{log}_2(6 + 2	ext{sqrt}(5)) 	imes 	ext{log}_2(6 - 2	ext{sqrt}(5))$.

Is there any identity for $	ext{log}_b(A) 	ext{log}_b(B)$?
Not a simple one.

Let's reconsider the structure $2^{5-x} + 2^{x-1} - 12 = 0$. 
Symmetry $f(x) = f(6-x)$. Roots $x_1, x_2$ are such that $x_1+x_2=6$. 
Product $x_1 x_2 = 9-d^2$. 

Let's check if $(6 	ext{ ± } 2	ext{sqrt}(5))$ are related to powers of 2 in a simple way.
Consider $(	ext{sqrt}(5)+1)^2 = 5 + 2	ext{sqrt}(5) + 1 = 6 + 2	ext{sqrt}(5)$.
So, $6 + 2	ext{sqrt}(5) = (rac{	ext{sqrt}(5)+1}{2} 	imes 2)^2 = (	ext{golden ratio} 	imes 2)^2$. 
$6 + 2	ext{sqrt}(5) = 4 	imes (rac{1+	ext{sqrt}(5)}{2})^2 = 4 	imes 	ext{golden ratio}^2$.

So, $2^{z_1} = w_1 = 4 	imes 	ext{golden ratio}^2$. 
$z_1 = 	ext{log}_2(4 	imes 	ext{golden ratio}^2) = 	ext{log}_2(4) + 	ext{log}_2(	ext{golden ratio}^2) = 2 + 2 	ext{log}_2(	ext{golden ratio})$.

And $2^{z_2} = w_2 = 6 - 2	ext{sqrt}(5)$. 
$(	ext{sqrt}(5)-1)^2 = 5 - 2	ext{sqrt}(5) + 1 = 6 - 2	ext{sqrt}(5)$.
So, $6 - 2	ext{sqrt}(5) = (rac{	ext{sqrt}(5)-1}{2} 	imes 2)^2 = (rac{	ext{golden ratio}^{-1} 	imes 2}{1})^2$. 
$6 - 2	ext{sqrt}(5) = 4 	imes (rac{	ext{sqrt}(5)-1}{2})^2 = 4 	imes (	ext{golden ratio}^{-1})^2 = 4 	imes (rac{1}{	ext{golden ratio}})^2$. 

$z_2 = 	ext{log}_2(4 	imes (rac{1}{	ext{golden ratio}})^2) = 	ext{log}_2(4) + 	ext{log}_2((rac{1}{	ext{golden ratio}})^2) = 2 + 2 	ext{log}_2(rac{1}{	ext{golden ratio}}) = 2 - 2 	ext{log}_2(	ext{golden ratio})$.

So, $z_1 = 2 + 2 	ext{log}_2(	ext{golden ratio})$ and $z_2 = 2 - 2 	ext{log}_2(	ext{golden ratio})$.

$z_1+z_2 = (2 + 2 	ext{log}_2(	ext{golden ratio})) + (2 - 2 	ext{log}_2(	ext{golden ratio})) = 4$. Correct.

$z_1 z_2 = (2 + 2 	ext{log}_2(	ext{golden ratio})) (2 - 2 	ext{log}_2(	ext{golden ratio}))$
$z_1 z_2 = 2^2 - (2 	ext{log}_2(	ext{golden ratio}))^2 = 4 - 4 (	ext{log}_2(	ext{golden ratio}))^2$. 

This still doesn't give a simple numerical value for $x_1 x_2$. 

Let's consider the possibility that the question is designed such that the final answer for $x_1 x_2$ is a simple integer. If this is the case, there might be a property I am missing, or a mistake in my derivations.

Could the question be from a context where $	ext{log}_2(	ext{golden ratio})$ is a known value that simplifies things?

Let's re-evaluate the original equation for symmetry one last time. $f(x) = 2^{5-x} + 2^{x-1} - 12 = 0$. We found $f(x) = f(6-x)$. This means roots are symmetric about $x=3$. So $x_1 = 3-d, x_2=3+d$, hence $x_1+x_2=6$ and $x_1x_2 = 9-d^2$. 

We had $2^{3-d} = y_1 = 12 + 4	ext{sqrt}(5)$ and $2^{3+d} = y_2 = 12 - 4	ext{sqrt}(5)$. (Wait, this assignment of $y_1, y_2$ might be arbitrary, it depends on which root is which). 

Let's use the relation $2^{x_1} = 12 + 4	ext{sqrt}(5)$ and $2^{x_2} = 12 - 4	ext{sqrt}(5)$.
We found $x_1 = 1 + 	ext{log}_2(6 + 2	ext{sqrt}(5))$ and $x_2 = 1 + 	ext{log}_2(6 - 2	ext{sqrt}(5))$.

Let $g = 	ext{log}_2(6 + 2	ext{sqrt}(5))$ and $h = 	ext{log}_2(6 - 2	ext{sqrt}(5))$.
$x_1 = 1+g$, $x_2 = 1+h$.
$x_1 x_2 = (1+g)(1+h) = 1 + g + h + gh$. 
We know $g+h = 	ext{log}_2((6+2	ext{sqrt}(5))(6-2	ext{sqrt}(5))) = 	ext{log}_2(16) = 4$.
So $x_1 x_2 = 1 + 4 + gh = 5 + gh$.

We need $gh = 	ext{log}_2(6 + 2	ext{sqrt}(5)) 	imes 	ext{log}_2(6 - 2	ext{sqrt}(5))$.

Is it possible that the question is designed such that $gh$ evaluates to a specific integer?

Let's reconsider the possibility of a typo. If the equation was $2^{x} + 2^{6-x} = 20$, we got $x_1x_2=8$. 
If the equation was $2^{2x} + 2^{2(3-x)} = c$, let $y=2^{2x}$. $y + 2^6/y = c$. 

Let's assume the intended answer is an integer. What integer could it be?
Possible values for $x_1x_2$ given $x_1+x_2=6$ and the roots are not symmetric around 3 (i.e., $x_1 
eq x_2$).
If $x_1=1, x_2=5$, then $x_1x_2=5$.
If $x_1=2, x_2=4$, then $x_1x_2=8$.
If $x_1=2.5, x_2=3.5$, then $x_1x_2=8.75$.

The calculation $x_1x_2 = 5 + gh$ where $g = 	ext{log}_2(6 + 2	ext{sqrt}(5))$ and $h = 	ext{log}_2(6 - 2	ext{sqrt}(5))$ needs to be resolved.

If $6 + 2	ext{sqrt}(5) = 2^k$ for some $k$, then $g=k$. And if $6 - 2	ext{sqrt}(5) = 2^m$, then $h=m$. 

We found $6 + 2	ext{sqrt}(5) = 4 (rac{1+	ext{sqrt}(5)}{2})^2 = 2^2 	imes 	ext{golden ratio}^2$. 
So $g = 	ext{log}_2(2^2 	imes 	ext{golden ratio}^2) = 2 + 2 	ext{log}_2(	ext{golden ratio})$.
And $h = 2 - 2 	ext{log}_2(	ext{golden ratio})$.

$gh = (2 + 2 	ext{log}_2(	ext{golden ratio})) (2 - 2 	ext{log}_2(	ext{golden ratio})) = 4 - 4 (	ext{log}_2(	ext{golden ratio}))^2$. 

So $x_1 x_2 = 5 + 4 - 4 (	ext{log}_2(	ext{golden ratio}))^2 = 9 - 4 (	ext{log}_2(	ext{golden ratio}))^2$. 
This is the same result as $9-d^2$ if $d=2 	ext{log}_2(	ext{golden ratio})$. 

Let's check if the source of the problem is from a specific curriculum or competition where such values are expected or can be simplified.

Assuming the problem is solvable and has a