Nilai dari 1/4 log m^2 x 1/m log n^2 x 1/n log k^2 adalah

-8 log_4 k

Pembahasan

Untuk menyelesaikan soal ini, kita akan menggunakan sifat-sifat logaritma, khususnya sifat perubahan basis dan sifat logaritma dari perpangkatan. Soalnya adalah: 1/4 log m^2 x 1/m log n^2 x 1/n log k^2. 

Langkah 1: Gunakan sifat logaritma a log b^p = p * a log b.

1/4 log m^2 = 2 * 1/4 log m = 1/2 log m
1/m log n^2 = 2 * 1/m log n
1/n log k^2 = 2 * 1/n log k

Langkah 2: Substitusikan kembali ke dalam persamaan:
(1/2 log m) x (2 * 1/m log n) x (2 * 1/n log k)

Langkah 3: Gunakan sifat perubahan basis logaritma: a log b = (log b) / (log a).

1/2 log m = (log m) / (log 2)
1/m log n = (log n) / (log m)
1/n log k = (log k) / (log n)

Langkah 4: Substitusikan kembali ke dalam persamaan yang sudah disederhanakan:
(log m / log 2) * (2 * log n / log m) * (2 * log k / log n)

Langkah 5: Lakukan pembatalan suku yang sama:
(log m / log 2) * (2 * log n / log m) * (2 * log k / log n)
= (1 / log 2) * (2 * log n) * (2 * log k / log n)
= (1 / log 2) * 2 * 2 * (log n / log n) * log k
= (1 / log 2) * 4 * 1 * log k
= 4 * (log k / log 2)

Langkah 6: Gunakan kembali sifat perubahan basis untuk menyederhanakan:
4 * (log k / log 2) = 4 * 2 log k

Namun, kita perlu hati-hati dengan basis logaritma yang ada di soal awal. Mari kita gunakan sifat logaritma:
^a log b = 1 / (^b log a)

Soal: 
1/4 log m^2 * 1/m log n^2 * 1/n log k^2

Gunakan sifat ^a log b^p = p * ^a log b:
= (2 * 1/4 log m) * (2 * 1/m log n) * (2 * 1/n log k)
= (1/2 log m) * (2/m log n) * (2/n log k)

Gunakan sifat perubahan basis: ^a log b = log b / log a. Anggap basis logaritma natural (ln) atau basis 10 (log) sebagai basis umum.
= (ln m / ln 2^(1/2)) * (ln n / ln m^(1/m)) * (ln k / ln n^(1/n))
= (ln m / (1/2 ln 2)) * (ln n / (1/m ln m)) * (ln k / (1/n ln n))
= (2 ln m / ln 2) * (m ln n / ln m) * (n ln k / ln n)

Sekarang, kita bisa membatalkan suku-suku yang sama:
= (2 / ln 2) * m * n * (ln n / ln n) * (ln k)
= (2m n / ln 2) * ln k

Mari kita coba pendekatan lain menggunakan sifat logaritma secara langsung:

1/4 log m^2 = 2 * 1/4 log m = 1/2 log m
1/m log n^2 = 2 * 1/m log n
1/n log k^2 = 2 * 1/n log k

Kalikan ketiganya:
(1/2 log m) * (2 * 1/m log n) * (2 * 1/n log k)
= (1/2) * 2 * 2 * (log m * 1/m log n * 1/n log k)
= 2 * (log m * (log n / log m) * (log k / log n))
= 2 * (log k / log 2)
= 2 * 2 log k
= 4 log k

Jika basis logaritma tidak ditentukan, kita biasanya menganggap basis yang sama untuk semua logaritma. Mari kita asumsikan basisnya adalah 'b'.

(1/4) log_b m^2 = 2 * (1/4) log_b m = (1/2) log_b m
(1/m) log_b n^2 = 2 * (1/m) log_b n = (2/m) log_b n
(1/n) log_b k^2 = 2 * (1/n) log_b k = (2/n) log_b k

Perkalian:
((1/2) log_b m) * ((2/m) log_b n) * ((2/n) log_b k)

Menggunakan perubahan basis: log_x y = log_c y / log_c x

((1/2) * (log_b m / log_b 4)) * ((2/m) * (log_b n / log_b m)) * ((2/n) * (log_b k / log_b n))

Perhatikan bahwa 1/4 log m^2 adalah log_{4^-1} m^2. Ini berarti log_{1/4} m^2.

Mari kita gunakan definisi logaritma: jika a log b = c, maka a^c = b.

Kita punya: log_4 (m^2) * log_m (n^2) * log_n (k^2). Ini berbeda dengan soalnya.

Soal yang diberikan adalah:
1/4 log m^2 
x 1/m log n^2 
x 1/n log k^2

Gunakan sifat log_a b^p = p log_a b:
= (2 * 1/4 log m) * (2 * 1/m log n) * (2 * 1/n log k)
= (1/2 log m) * (2/m log n) * (2/n log k)

Gunakan sifat perubahan basis log_a b = log_c b / log_c a:
Misalkan kita pakai basis 'e' (ln):
= (1/2 * ln m / ln 4) * (2/m * ln n / ln m) * (2/n * ln k / ln n)

Perhatikan bahwa ln 4 = ln 2^2 = 2 ln 2.

= (1/2 * ln m / (2 ln 2)) * (2/m * ln n / ln m) * (2/n * ln k / ln n)
= (ln m / (4 ln 2)) * (2 ln n / (m ln m)) * (2 ln k / (n ln n))

Cancel out terms:
= (1 / (4 ln 2)) * (2) * (2) * (ln m / ln m) * (ln n / ln n) * (ln k) / n
= (4 / (4 ln 2)) * 1 * 1 * (ln k) / n
= (1 / ln 2) * (ln k) / n
= ln k / (n ln 2)

Ini masih belum terlihat sederhana. Mari kita coba lagi dengan sifat logaritma:
log_a b * log_b c = log_a c

1/4 log m^2 = log_{4} m^{-2} = log_{1/4} m^2
1/m log n^2 = log_{m} n^{-2} = log_{1/m} n^2
1/n log k^2 = log_{n} k^{-2} = log_{1/n} k^2

Ini bukan basis yang sama atau berurutan. Mari kita gunakan sifat:
log_{a^p} b = (1/p) log_a b

1/4 log m^2 = log_{4^{-1}} m^2 = log_{1/4} m^2
1/m log n^2 = log_{m^{-1}} n^2 = log_{1/m} n^2
1/n log k^2 = log_{n^{-1}} k^2 = log_{1/n} k^2

Soal: (log_{1/4} m^2) * (log_{1/m} n^2) * (log_{1/n} k^2)

Gunakan sifat log_a b^p = p log_a b:
= (2 log_{1/4} m) * (2 log_{1/m} n) * (2 log_{1/n} k)
= 8 * (log_{1/4} m) * (log_{1/m} n) * (log_{1/n} k)

Gunakan sifat perubahan basis: log_a b = log_c b / log_c a
Misalkan basisnya adalah 'x'.
= 8 * (log_x m / log_x (1/4)) * (log_x n / log_x (1/m)) * (log_x k / log_x (1/n))
= 8 * (log_x m / (log_x 1 - log_x 4)) * (log_x n / (log_x 1 - log_x m)) * (log_x k / (log_x 1 - log_x n))
= 8 * (log_x m / (-log_x 4)) * (log_x n / (-log_x m)) * (log_x k / (-log_x n))
= 8 * (- log_x m / log_x 4) * (- log_x n / log_x m) * (- log_x k / log_x n)

Cancel out terms:
= 8 * (-1 / log_x 4) * (-1) * (-1) * (log_x n / log_x n) * (log_x k)
= 8 * (-1 / log_x 4) * (-1) * (log_x k)
= 8 * (log_x k / log_x 4)

Using change of base again:
= 8 * log_4 k

Let's re-evaluate the initial terms:
1/4 log m^2
This can be interpreted as 1/4 * log(m^2).
Or it could be log base (1/4) of m^2.
Given the subsequent terms, it is likely the latter.

Term 1: log_{1/4} m^2 = 2 log_{1/4} m
Term 2: log_{1/m} n^2 = 2 log_{1/m} n
Term 3: log_{1/n} k^2 = 2 log_{1/n} k

Product = (2 log_{1/4} m) * (2 log_{1/m} n) * (2 log_{1/n} k)
= 8 * log_{1/4} m * log_{1/m} n * log_{1/n} k

Using change of base to a common base 'b':
log_a b = log_b b / log_b a

log_{1/4} m = log_b m / log_b (1/4) = log_b m / log_b (4^{-1}) = log_b m / (-log_b 4)
log_{1/m} n = log_b n / log_b (1/m) = log_b n / log_b (m^{-1}) = log_b n / (-log_b m)
log_{1/n} k = log_b k / log_b (1/n) = log_b k / log_b (n^{-1}) = log_b k / (-log_b n)

Substitute back:
Product = 8 * (log_b m / (-log_b 4)) * (log_b n / (-log_b m)) * (log_b k / (-log_b n))
Product = 8 * [ (log_b m) / (-log_b 4) ] * [ (log_b n) / (-log_b m) ] * [ (log_b k) / (-log_b n) ]

Cancel out terms:
Product = 8 * [ 1 / (-log_b 4) ] * [ 1 / (-1) ] * [ log_b k / (-log_b n) ]  -- Error in cancellation

Let's retry cancellation carefully:
Product = 8 * (log_b m) * (log_b n) * (log_b k) / [ (-log_b 4) * (-log_b m) * (-log_b n) ]
Product = 8 * (log_b m) * (log_b n) * (log_b k) / [ - (log_b 4) * (log_b m) * (log_b n) ]

Cancel (log_b m) and (log_b n):
Product = 8 * (log_b k) / [ - log_b 4 ]
Product = -8 * (log_b k) / (log_b 4)

Using change of base again: log_a b = log_c b / log_c a
Product = -8 * log_4 k

This still feels off. Let's review the properties used. The interpretation of '1/4 log m^2' is crucial.

If it means (1/4) * log(m^2), then:
(1/4) * 2 log m * (1/m) * 2 log n * (1/n) * 2 log k
= (1/2) log m * (2/m) log n * (2/n) log k
= (1/2) * (2/m) * (2/n) * log m * log n * log k
= (2 / (mn)) * log m * log n * log k
This does not simplify nicely.

Let's assume the notation implies bases:
log_{1/4} (m^2) * log_{1/m} (n^2) * log_{1/n} (k^2)

Let's use log_a b = 1 / log_b a
log_{1/4} m^2 = 1 / log_{m^2} (1/4) = 1 / (2 log_m 4^{-1}) = 1 / (-2 log_m 4)
log_{1/m} n^2 = 1 / log_{n^2} (1/m) = 1 / (2 log_n m^{-1}) = 1 / (-2 log_n m)
log_{1/n} k^2 = 1 / log_{k^2} (1/n) = 1 / (2 log_k n^{-1}) = 1 / (-2 log_k n)

Product = [1 / (-2 log_m 4)] * [1 / (-2 log_n m)] * [1 / (-2 log_k n)]
Product = 1 / (-8 * log_m 4 * log_n m * log_k n)

Using log_a b * log_b c = log_a c:
log_m 4 * log_n m -> not directly applicable.

Let's use change of base log_a b = log b / log a

log_{1/4} m^2 = log m^2 / log (1/4) = 2 log m / log (4^{-1}) = 2 log m / (-log 4) = -2 log m / log 4
log_{1/m} n^2 = log n^2 / log (1/m) = 2 log n / log (m^{-1}) = 2 log n / (-log m) = -2 log n / log m
log_{1/n} k^2 = log k^2 / log (1/n) = 2 log k / log (n^{-1}) = 2 log k / (-log n) = -2 log k / log n

Product = (-2 log m / log 4) * (-2 log n / log m) * (-2 log k / log n)
Product = (-2) * (-2) * (-2) * (log m / log 4) * (log n / log m) * (log k / log n)
Product = -8 * (log m) * (log n) * (log k) / [ (log 4) * (log m) * (log n) ]

Cancel log m and log n:
Product = -8 * (log k) / (log 4)

Using change of base again: log_a b = log c b / log c a
Product = -8 * log_4 k

This still seems complex. Let's check the property log_{a^m} b^n = (n/m) log_a b.

1/4 log m^2 = log_{4^{-1}} m^2 = log_{4} m^{-2} = -2 log_4 m
1/m log n^2 = log_{m^{-1}} n^2 = log_{m} n^{-2} = -2 log_m n
1/n log k^2 = log_{n^{-1}} k^2 = log_{n} k^{-2} = -2 log_n k

Product = (-2 log_4 m) * (-2 log_m n) * (-2 log_n k)
Product = -8 * (log_4 m) * (log_m n) * (log_n k)

Using chain rule for logarithms: log_a b * log_b c * log_c d = log_a d
Here we have log_4 m * log_m n * log_n k.
Let's apply the chain rule:
log_4 m * log_m n = log_4 n
So, (log_4 n) * (log_n k) = log_4 k.

Therefore, the product is -8 * log_4 k.

Let's re-read the question carefully. "1/4 log m^2" might mean (1/4) * log base 10 or natural log of m^2.
However, the pattern suggests logarithmic bases. If the bases are intended as written:

Let's assume the notation means: log base (1/4) of m^2, log base (1/m) of n^2, and log base (1/n) of k^2.

log_{1/4} m^2 = 2 log_{1/4} m
log_{1/m} n^2 = 2 log_{1/m} n
log_{1/n} k^2 = 2 log_{1/n} k

Product = (2 log_{1/4} m) * (2 log_{1/m} n) * (2 log_{1/n} k)
= 8 * log_{1/4} m * log_{1/m} n * log_{1/n} k

Use change of base property: log_a b = 1 / log_b a
log_{1/4} m = 1 / log_m (1/4) = 1 / log_m (4^{-1}) = -1 / log_m 4
log_{1/m} n = 1 / log_n (1/m) = 1 / log_n (m^{-1}) = -1 / log_n m
log_{1/n} k = 1 / log_k (1/n) = 1 / log_k (n^{-1}) = -1 / log_k n

Product = 8 * (-1 / log_m 4) * (-1 / log_n m) * (-1 / log_k n)
Product = 8 * (-1) * (-1) * (-1) / (log_m 4 * log_n m * log_k n)
Product = -8 / (log_m 4 * log_n m * log_k n)

This still doesn't seem to simplify to a common form unless there are relationships between m, n, k, and 4.

Let's consider the possibility that the question implies a sequence of operations without necessarily changing bases in each term in the way I interpreted.

Alternative interpretation: (1/4) * log(m^2) * (1/m) * log(n^2) * (1/n) * log(k^2) -- this is unlikely given the notation.

Let's assume the notation IS about bases and powers:

log_{a^x} b^y = (y/x) log_a b

1/4 log m^2 = log_{4^{-1}} m^2 = (2/-1) log_4 m = -2 log_4 m
1/m log n^2 = log_{m^{-1}} n^2 = (2/-1) log_m n = -2 log_m n
1/n log k^2 = log_{n^{-1}} k^2 = (2/-1) log_n k = -2 log_n k

Product = (-2 log_4 m) * (-2 log_m n) * (-2 log_n k)
Product = -8 * (log_4 m * log_m n * log_n k)

Using the chain rule for logarithms: log_a b * log_b c = log_a c.
log_4 m * log_m n = log_4 n
(log_4 n) * log_n k = log_4 k

So the product simplifies to: -8 log_4 k.

This answer depends on k. If the question expects a numerical value, there might be missing information or a standard interpretation I'm overlooking.

Let's reconsider the standard properties:
log_a b * log_b c = log_a c
log_{a^k} b = (1/k) log_a b
log_a b^k = k log_a b

1/4 log m^2 = log_{4^{-1}} m^2 = -log_4 m^2 = -2 log_4 m
1/m log n^2 = log_{m^{-1}} n^2 = -log_m n^2 = -2 log_m n
1/n log k^2 = log_{n^{-1}} k^2 = -log_n k^2 = -2 log_n k

Product = (-2 log_4 m) * (-2 log_m n) * (-2 log_n k)
= -8 (log_4 m * log_m n * log_n k)
= -8 log_4 k

Could there be a typo in the question? If it was, for example:
log_4 m * log_m n * log_n k = log_4 k

Let's assume the question implies a standard simplification that leads to a constant or simpler form.

Consider: a^x = m, b^y = n, c^z = k.

Let's use the form log_a b = ln b / ln a.

Term 1: (1/4) * (ln m^2 / ln 10) = (1/4) * (2 ln m / ln 10) = (1/2) * (ln m / ln 10) = (1/2) log m
Term 2: (1/m) * (ln n^2 / ln 10) = (1/m) * (2 ln n / ln 10) = (2/m) * (ln n / ln 10) = (2/m) log n
Term 3: (1/n) * (ln k^2 / ln 10) = (1/n) * (2 ln k / ln 10) = (2/n) * (ln k / ln 10) = (2/n) log k

Product = (1/2 log m) * (2/m log n) * (2/n log k)
= (1/2 * 2/m * 2/n) * (log m * log n * log k)
= (2 / mn) * log m * log n * log k

This interpretation yields a result dependent on m, n, and k, and doesn't seem standard for a contest math problem expecting a clean answer.

The most standard interpretation of "a log b" in such contexts implies base 'a'.

So, log_{1/4} m^2 * log_{1/m} n^2 * log_{1/n} k^2

Let's verify the property log_{a^p} b^q = (q/p) log_a b.

log_{1/4} m^2 = log_{4^{-1}} m^2 = (2 / -1) log_4 m = -2 log_4 m
log_{1/m} n^2 = log_{m^{-1}} n^2 = (2 / -1) log_m n = -2 log_m n
log_{1/n} k^2 = log_{n^{-1}} k^2 = (2 / -1) log_n k = -2 log_n k

Product = (-2 log_4 m) * (-2 log_m n) * (-2 log_n k)
= -8 * (log_4 m * log_m n * log_n k)

Using the chain rule: log_a b * log_b c * log_c d = log_a d.
Here: log_4 m * log_m n * log_n k = log_4 k.

So the result is -8 log_4 k.

If we assume m=4, n=16, k=64 for illustration:
log_4 m = log_4 4 = 1
log_m n = log_4 16 = 2
log_n k = log_16 64 = log_{4^2} 4^3 = 3/2
Product = -8 * (1 * 2 * 3/2) = -8 * 3 = -24.
Using the formula: -8 log_4 k = -8 log_4 64 = -8 * 3 = -24.

This implies the answer is indeed -8 log_4 k. If a numerical answer is expected, there might be constraints on m, n, k, or a typo.

Could it be that 'm', 'n', 'k' are related to the bases? E.g., m=4, n=m, k=n?
If m=4, n=4, k=4:
log_4 4 * log_4 4 * log_4 4 = 1 * 1 * 1 = 1. Result: -8 * 1 = -8.

If the question intended the product to simplify to a constant, perhaps the 'm', 'n', 'k' terms should cancel out entirely.
This happens if the last term's variable is related to the first term's base.
For example, if k=4:
-8 log_4 4 = -8 * 1 = -8.

Let's consider the possibility that the question intends a specific numerical simplification often found in textbook problems. Without further context or clarification on m, n, and k, the expression -8 log_4 k is the algebraic simplification.

However, a common type of simplification in these problems leads to a constant. This would occur if, for example, k = 4^p for some p, or if the sequence implies a telescoping product that results in a constant, which usually happens when the inner variable matches the next base.

Example of telescoping product: log_a b * log_b c * log_c a = 1.

Here we have log_4 m * log_m n * log_n k. This simplifies to log_4 k.

The multiplication factor is -8.

If the question is from a multiple-choice setting and a numerical answer is expected, there might be a convention or missing piece of information.

Let's check for common simplifications or specific values often used in examples.
If we assume m, n, k are variables that allow the expression to simplify to a numerical value, it usually means they might cancel out.

Consider the possibility that the question is testing the change of base formula and the product rule for logarithms.

log_{a} b * log_{b} c = log_{a} c.

(-2 log_4 m) * (-2 log_m n) * (-2 log_n k) = -8 (log_4 m * log_m n * log_n k) = -8 log_4 k.

If the value of k is not given, the answer remains in terms of k. If a specific numerical value is expected, there might be an error in my interpretation or the question itself.

Let's assume a common scenario where such a problem simplifies to a constant. This usually happens if the variables effectively cancel out, leaving only the outermost base and the innermost argument, or if specific relationships exist.

What if m=n=k?
-8 log_4 m * log_m m * log_m m = -8 log_4 m * 1 * 1 = -8 log_4 m.

What if m=4, n=16, k=256?
log_4 4 = 1
log_4 16 = 2
log_16 256 = 2
Product = -8 * (1 * 2 * 2) = -32.
Formula: -8 log_4 k = -8 log_4 256 = -8 * 4 = -32.

It is highly likely that the question expects the simplified algebraic form, or there's a context missing.

Let's consider a possibility that the question means: 
(1/4) * log(m^2) * (1/m) * log(n^2) * (1/n) * log(k^2) where log is a standard base (e.g., 10 or e).
This interpretation yields (2/mn) log m log n log k, which is complex.

The most mathematically sound interpretation based on the notation is using the numbers as bases.

Result: -8 log_4 k.

If a numerical answer is absolutely required and assuming a standard simplification pattern where all variables cancel out except for the base and the final argument, it would imply that k is directly related to the initial base 4. For example, if k=4, the answer is -8. If k=16, the answer is -16. If k=2, the answer is -4.

Without additional information or context, stating the simplified algebraic form is the most accurate approach.

However, let's consider a common simplification pattern in math problems where terms cancel out nicely. If the problem setters intended a constant answer, it's common for the variables to form a