Sets, Relations & Functions
General
Grade 11

Question:

<p>Let <span class="math-inline">\(f_1(x) = 2^{f_2(x)}\)</span>, <span class="math-inline">\(f_2(x) = 2012^{f_3(x)}\)</span>, <span class="math-inline">\(f_3(x) = \left(\frac{1}{2013}\right)^{f_4(x)}\)</span>, where <span class="math-inline">\(f_4(x) = \log_{2013}(\log_x 2012)\)</span>. Find the range of <span class="math-inline">\(f_1(x)\)</span>.</p>

Step-by-Step Solution

Key Concept: General
<div class="solution"><p><strong>Key Idea:</strong> Collapse the tower using <span class="math-inline">\(a^{\log_a t} = t\)</span> and reciprocal log identity.</p><p><strong>Step 1:</strong> Domain: <span class="math-inline">\(\log_x 2012 > 0\)</span> requires <span class="math-inline">\(x > 1\)</span>.</p><p><strong>Step 2:</strong> <span class="math-block">\[f_3(x) = \left(\frac{1}{2013}\right)^{\log_{2013}(\log_x 2012)} = \frac{1}{\log_x 2012} = \log_{2012} x\]</span></p><p><strong>Step 3:</strong> <span class="math-inline">\(f_2(x) = 2012^{\log_{2012} x} = x\)</span></p><p><strong>Step 4:</strong> <span class="math-inline">\(f_1(x) = 2^x\)</span>, <span class="math-inline">\(x > 1\)</span>, so range is <span class="math-inline">\((2,\infty)\)</span></p><p><strong>Answer: <span class="math-inline">\((2,\infty)\)</span></strong></p><div class="trap-box"><strong>Trap:</strong> Domain is critical — <span class="math-inline">\(\log_x 2012\)</span> must also be positive (not just defined) because it is an argument of another log.</div><div class="key-concept"><strong>Key Concept:</strong> Nested exponent-log telescoping using <span class="math-inline">\(a^{\log_a t}=t\)</span></div></div>
Correct Answer: (2, ∞)

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