Question 5 of 14intermediate🔧 ApplyLong Answer4 marks

Prove that 312×314×318…=33^{\frac{1}{2}} \times 3^{\frac{1}{4}} \times 3^{\frac{1}{8}} \ldots = 3

Correct Answer

To prove:

312×314×318…=33^{\frac{1}{2}} \times 3^{\frac{1}{4}} \times 3^{\frac{1}{8}} \ldots = 3

Step 1: Using the law of exponents am×an=am+na^m \times a^n = a^{m+n}, we can combine all terms with base 3:

312×314×318…=312+14+18+…3^{\frac{1}{2}} \times 3^{\frac{1}{4}} \times 3^{\frac{1}{8}} \ldots = 3^{\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots}

Step 2: The exponent is an infinite G.P. of the form 12,14,18,…\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \ldots with first term a=12a = \frac{1}{2} and common ratio r=12r = \frac{1}{2}.

Step 3: For an infinite G.P. where ∣r∣<1|r| < 1, as n→∞n \to \infty, we have rn→0r^n \to 0. The sum to infinity is given by:

S∞=a1−r=121−12=1212=1S_\infty = \frac{a}{1-r} = \frac{\frac{1}{2}}{1-\frac{1}{2}} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1

Step 4: Therefore, the original expression becomes:

312+14+18+…=31=33^{\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots} = 3^1 = 3

∴ Hence proved.

Exercise: Exercise 8.3 | Q: 5 | (Page Page 2)
For More Understanding

Explanation

This question tests the student's understanding of laws of exponents combined with the concept of infinite geometric progression. The key insight is recognizing that when multiplying powers with the same base, the exponents add up.

The sum of the infinite G.P. in the exponent equals 1, making the final result equal to 31=33^1 = 3.

The textbook context provides the foundation for infinite G.P. sums, showing that as n→∞n \to \infty, rn→0r^n \to 0 when ∣r∣<1|r| < 1.

Solution Steps

  1. Step 1: Apply law of exponents to combine terms: product becomes 3 raised to sum of exponents

  2. Step 2: Identify the exponent series as infinite G.P. with a = 1/2, r = 1/2

  3. Step 3: Apply infinite G.P. sum formula: S = a/(1-r) = 1

  4. Step 4: Substitute back to get 313^{1} = 3

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