Indices and Surds: IPMAT practice questions

Solution

2^{x + 1} = 2 × 2^x, so the left side is (2 + 1) × 2^x = 3 × 2^x.

2^x = 24 / 3 = 8 = 2³, so x = 3.

Solution

Raise each to the power 12 (the LCM of the root orders) so all become whole numbers.

√3 → 3⁶ = 729; 3¹⁴ → 3³ = 27; 12¹⁶ → 12² = 144; √2 → 2⁶ = 64.

The largest is √3.

Solution

Rationalise each term: 1/(√k + √(k+1)) = √(k+1) − √k.

The sum telescopes: (√2 − √1) + (√3 − √2) + … + (√25 − √24) = √25 − 1.

√25 − 1 = 5 − 1 = 4.

Solution

Look for x + y = 29 and xy = 120.

x = 24 and y = 5 work, since (√24 + √5)² = 24 + 5 + 2√120.

So the value is √24 + √5.

Solution

2^{x + 3} = 8 × 2^x, so the left side is (8 + 1) × 2^x = 9 × 2^x.

2^x = 36 / 9 = 4 = 2², so x = 2.

Solution

Look for x + y = 24 and xy = 108.

x = 18 and y = 6 work, since (√18 + √6)² = 18 + 6 + 2√108.

So the value is √18 + √6.

Solution

Raise each to the power 60 (the LCM of the root orders) so all become whole numbers.

4¹⁵ → 4¹² = 16777216; √2 → 2³⁰ = 1073741824; 30¹¹² → 30⁵ = 24300000; √3 → 3³⁰ = 205891132094649.

The largest is √3.

Solution

Look for x + y = 29 and xy = 54.

x = 27 and y = 2 work, since (√27 + √2)² = 27 + 2 + 2√54.

So the value is √27 + √2.

Solution

(4 + √15)(4 − √15) = 16 − 15 = 1, so 1/x = 4 − √15.

x + 1/x = 8.

x³ + 1/x³ = (x + 1/x)³ − 3(x + 1/x) = 512 − 24 = 488.

Solution

Look for x + y = 38 and xy = 192.

x = 32 and y = 6 work, since (√32 + √6)² = 32 + 6 + 2√192.

So the value is √32 + √6.

Solution

81 = 3⁴, so 81²⁴ = 3² = 9.

27 = 3³, so 27⁴³ = 3⁴ = 81.

Result: 9 ÷ 81 = 19.

Solution

Let each equal k. Then 2 = k^1/x, 7 = k^1/y and 14 = k^1/z.

Since 2 × 7 = 14: k^{1/x + 1/y} = k^1/z, so 1/x + 1/y = 1/z.

That rearranges to z = xy/(x + y).

Solution

125 = 5³, so 125²³ = 5² = 25.

16 = 2⁴, so 16⁴⁴ = 2⁴ = 16.

Result: 25 × 16 = 400.

Solution

Raise each to the power 12 (the LCM of the root orders) so all become whole numbers.

12¹⁶ → 12² = 144; 30¹¹² → 30¹ = 30; √3 → 3⁶ = 729; 3¹⁴ → 3³ = 27.

The largest is √3.

Solution

(2 + √3)(2 − √3) = 4 − 3 = 1, so 1/x = 2 − √3.

x + 1/x = 4.

Solution

3^{x + 3} = 27 × 3^x, so the left side is (27 + 1) × 3^x = 28 × 3^x.

3^x = 20412 / 28 = 729 = 3⁶, so x = 6.

Solution

Rationalise each term: 1/(√k + √(k+1)) = √(k+1) − √k.

The sum telescopes: (√2 − √1) + (√3 − √2) + … + (√16 − √15) = √16 − 1.

√16 − 1 = 4 − 1 = 3.

Solution

8 = 2³, so 8²³ = 2² = 4.

16 = 2⁴, so 16¹⁴ = 2¹ = 2.

Result: 4 × 2 = 8.

Solution

Raise each to the power 12 (the LCM of the root orders) so all become whole numbers.

√2 → 2⁶ = 64; 3¹³ → 3⁴ = 81; 30¹¹² → 30¹ = 30; 5¹⁴ → 5³ = 125.

The largest is 5¹⁴.

Solution

Rationalise each term: 1/(√k + √(k+1)) = √(k+1) − √k.

The sum telescopes: (√2 − √1) + (√3 − √2) + … + (√49 − √48) = √49 − 1.

√49 − 1 = 7 − 1 = 6.

Solution

(5 + √24)(5 − √24) = 25 − 24 = 1, so 1/x = 5 − √24.

x + 1/x = 10.

x³ + 1/x³ = (x + 1/x)³ − 3(x + 1/x) = 1000 − 30 = 970.

Solution

Let each equal k. Then 3 = k^1/x, 5 = k^1/y and 15 = k^1/z.

Since 3 × 5 = 15: k^{1/x + 1/y} = k^1/z, so 1/x + 1/y = 1/z.

That rearranges to z = xy/(x + y).

Solution

Let each equal k. Then 3 = k^1/x, 7 = k^1/y and 21 = k^1/z.

Since 3 × 7 = 21: k^{1/x + 1/y} = k^1/z, so 1/x + 1/y = 1/z.

That rearranges to z = xy/(x + y).

Solution

3^{x + 2} = 9 × 3^x, so the left side is (9 + 1) × 3^x = 10 × 3^x.

3^x = 2430 / 10 = 243 = 3⁵, so x = 5.