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The question asks to simplify logarithmic expressions. Let's simplify each part step by step.

### a. \( ^6\log 3 + ^6\log 12 \)
We can use the logarithmic property:
\[
^a\log x + ^a\log y = ^a\log(x \cdot y)
\]
Thus, this becomes:
\[
^6\log (3 \times 12) = ^6\log 36
\]

### b. \( ^6\log 9 + 2 ^6\log 2 - 2 ^6\log 6 \)
We simplify step by step using properties of logarithms:
\[
2^a\log b = ^a\log (b^2)
\]
So the expression becomes:
\[
^6\log 9 + ^6\log 2^2 - ^6\log 6^2
\]
\[
= ^6\log 9 + ^6\log 4 - ^6\log 36
\]
Now, using the property:
\[
^a\log x + ^a\log y - ^a\log z = ^a\log\left(\frac{x \cdot y}{z}\right)
\]
We have:
\[
^6\log\left(\frac{9 \times 4}{36}\right) = ^6\log 1 = 0
\]

### c. \( ^2\log 12 + ^4\log \frac{16}{9} \)
Using logarithmic property:
\[
^a\log \frac{x}{y} = ^a\log x - ^a\log y
\]
We handle the second term first:
\[
^4\log \frac{16}{9} = ^4\log 16 - ^4\log 9 = \frac{\log 16}{\log 4} - \frac{\log 9}{\log 4} = \frac{4 \log 2}{2 \log 2} - \frac{2 \log 3}{2 \log 2} = 0 
\]


Sederhanakan soal-soal di bawah ini: 
 a. 6log3 + 6log12 
 b. 6log9 + 26log2 − 26log6 
 c. 2log12 + 4log16/9 
 d. 5log3 × 3log1/25 
 e. 2log10 × 6log4 × log216

Simplify logarithmic expressions involving products, quotients, and powers. Learn how to apply the product, quotient, and power rules for logarithms with detailed examples, including questions such as 6log3 + 6log12 and more. Suitable for high school students (Grades 10-12).

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