If $\log_2 [3 + \log_3 \{4 + \log_4 (x - 1)\}] - 2 = 0$, then $4x$ equals

Given: $\log_2 [3 + \log_3 \{4 + \log_4 (x - 1)\}] - 2 = 0$

Find: The numerical value of $4x$

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Understanding Logarithms:

If $\log_a (b) = c$, then $a^c = b$

This means: "To what power must we raise $a$ to get $b$? The answer is $c$."

For example: $\log_2 (8) = 3$ because $2^3 = 8$

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Starting with: $\log_2 [3 + \log_3 \{4 + \log_4 (x - 1)\}] - 2 = 0$

$\log_2 [3 + \log_3 \{4 + \log_4 (x - 1)\}] = 2$

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Since $\log_2 (\text{something}) = 2$, we know that $2^2 = \text{something}$

Therefore: $3 + \log_3 \{4 + \log_4 (x - 1)\} = 2^2 = 4$

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$3 + \log_3 \{4 + \log_4 (x - 1)\} = 4$

$\log_3 \{4 + \log_4 (x - 1)\} = 1$

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Since $\log_3 (\text{something}) = 1$, we know that $3^1 = \text{something}$

Therefore: $4 + \log_4 (x - 1) = 3^1 = 3$

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$4 + \log_4 (x - 1) = 3$

$\log_4 (x - 1) = -1$

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Since $\log_4 (\text{something}) = -1$, we know that $4^{-1} = \text{something}$

Remember: $4^{-1} = \tfrac{1}{4}$ (negative exponent means reciprocal)

Therefore: $x - 1 = \tfrac{1}{4}$

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$x - 1 = \tfrac{1}{4}$

$x = 1 + \tfrac{1}{4} = \tfrac{4}{4} + \tfrac{1}{4} = \tfrac{5}{4}$

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$4x = 4 \times \tfrac{5}{4} = 5$

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Answer: $4x = 5$

Key Takeaway: When we solve nested logarithm equations, we work from the outside in, converting each logarithm using the fundamental property $\log_a (b) = c \rightarrow a^c = b$. This systematic approach prevents errors and makes complex problems manageable.