If $\log (2^a \times 3^b \times 5^c)$ is the arithmetic mean of $\log (2^2 \times 3^3 \times 5)$, $\log (2^6 \times 3 \times 5^7)$, and $\log (2 \times 3^2 \times 5^4)$, then a equals

We need to find the value of 'a' when $\log (2^a \times 3^b \times 5^c)$ equals the arithmetic mean of three logarithmic expressions.

The arithmetic mean of three numbers is their sum divided by 3.

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Since $\log (2^a \times 3^b \times 5^c)$ is the arithmetic mean of the three given expressions:

$\log (2^a \times 3^b \times 5^c) = \frac{1}{3}[\log (2^2 \times 3^3 \times 5) + \log (2^6 \times 3 \times 5^7) + \log (2 \times 3^2 \times 5^4)]$

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Using the property $\log(A) + \log(B) = \log(A \times B)$:

$\frac{1}{3}[\log (2^2 \times 3^3 \times 5) + \log (2^6 \times 3 \times 5^7) + \log (2 \times 3^2 \times 5^4)]$

$= \frac{1}{3} \log[(2^2 \times 3^3 \times 5) \times (2^6 \times 3 \times 5^7) \times (2 \times 3^2 \times 5^4)]$

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Combining like bases:

For powers of 2: $2^2 \times 2^6 \times 2^1 = 2^{2+6+1} = 2^9$

For powers of 3: $3^3 \times 3^1 \times 3^2 = 3^{3+1+2} = 3^6$

For powers of 5: $5^1 \times 5^7 \times 5^4 = 5^{1+7+4} = 5^{12}$

Our equation becomes:

$\log (2^a \times 3^b \times 5^c) = \frac{1}{3} \log(2^9 \times 3^6 \times 5^{12})$

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Using $\frac{1}{3} \log(X) = \log(X^{1/3})$:

$\log (2^a \times 3^b \times 5^c) = \log[(2^9 \times 3^6 \times 5^{12})^{1/3}]$

Using $(A^m)^n = A^{mn}$:

$\log (2^a \times 3^b \times 5^c) = \log(2^{9/3} \times 3^{6/3} \times 5^{12/3})$

$\log (2^a \times 3^b \times 5^c) = \log(2^3 \times 3^2 \times 5^4)$

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Since logarithm is a one-to-one function, if $\log(X) = \log(Y)$, then $X = Y$.

Therefore:

$2^a \times 3^b \times 5^c = 2^3 \times 3^2 \times 5^4$

Comparing the powers of each prime factor:

Power of 2: $a = 3$

Power of 3: $b = 2$

Power of 5: $c = 4$

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Therefore, $a = 3$.

When dealing with logarithms of products, we use the property $\log(A) + \log(B) = \log(A \times B)$ to combine terms, then equate the arguments when the logarithms are equal.