Basics & LogarithmshardFree

Basics & Logarithms: Find Value

JEE Maths question with a full step-by-step solution.

Question
Find the value of
log⁡5((55+5)(55+5)2(55+5)3⋯(6+25)215−18215−18215−18⋯3333)\log_{\sqrt5}\left(\frac{\sqrt{\left(5\sqrt5+5\right)\sqrt{\left(5\sqrt5+5\right)^2\sqrt{\left(5\sqrt5+5\right)^3\sqrt{\cdots}}}}} {\left(6+2\sqrt5\right)\sqrt[3]{215-18\sqrt[3]{215-18\sqrt[3]{215-18\sqrt[3]{\cdots}}}}}\right)
Solution
Answer: 2
Step 1: Simplifying the numerator by collecting the exponents of a=55+5a = 5\sqrt5+5.
N=aa2a3⋯=a1/2⋅(a2)1/4⋅(a3)1/8⋯=a E,N = \sqrt{a\sqrt{a^2\sqrt{a^3\sqrt{\cdots}}}} = a^{1/2}\cdot\left(a^2\right)^{1/4}\cdot\left(a^3\right)^{1/8}\cdots = a^{\,E},
 where
E=12+24+38+⋯=∑k=1∞k2k.E = \frac12+\frac{2}{4}+\frac{3}{8}+\cdots = \sum_{k=1}^{\infty}\frac{k}{2^{k}}.
 Step 2: Sum that series. Using ∑k≥1kxk=x(1−x)2\displaystyle\sum_{k\ge1}kx^{k} = \frac{x}{(1-x)^2} with x=12x = \tfrac12,
E=1/2(1/2)2=2.E = \frac{1/2}{\left(1/2\right)^2} = 2 .
 So N=a2N = a^2. Step 3: Simplify a2a^2.
a=55+5=5(5+1),a = 5\sqrt5+5 = 5\left(\sqrt5+1\right),
a2=25(5+1)2=25(5+25+1)=25(6+25).a^2 = 25\left(\sqrt5+1\right)^2 = 25\left(5+2\sqrt5+1\right) = 25\left(6+2\sqrt5\right).
 Step 4: Handle the nested cube root by self-similarity. Let
y=215−18215−18⋯333.y = \sqrt[3]{215-18\sqrt[3]{215-18\sqrt[3]{\cdots}}}.
 The expression inside repeats itself, so y=215−18y3y = \sqrt[3]{215-18y}, i.e.
y3+18y−215=0.y^3+18y-215 = 0 .
 Step 5: Solve the cubic. Try y=5y = 5: 125+90−215=0125+90-215 = 0 . Factoring,
y3+18y−215=(y−5)(y2+5y+43),y^3+18y-215 = \left(y-5\right)\left(y^2+5y+43\right),
 and y2+5y+43y^2+5y+43 has discriminant 25−172<025-172 < 0, so y=5y = 5 is the only real value. Step 6: Assemble the fraction.
N(6+25)y=25(6+25)5(6+25)=5.\frac{N}{\left(6+2\sqrt5\right)y} = \frac{25\left(6+2\sqrt5\right)}{5\left(6+2\sqrt5\right)} = 5 .
 The surd 6+256+2\sqrt5 cancels exactly. Step 7: Take the logarithm.
log⁡55=log⁡5(5)2=2.\log_{\sqrt5}5 = \log_{\sqrt5}\left(\sqrt5\right)^2 = 2 .
 Answer: 22
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