4^(2n+3) = 8^(n+5)? - Testpoint

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4^(2n+3) = 8^(n+5)?

9
7
8
None of these

Explanation

Let's solve the equation:

4^(2n+3) = 8^(n+5)

Expressing both sides with base 2:

(2^2)^(2n+3) = (2^3)^(n+5)

2^(4n+6) = 2^(3n+15)

Equating exponents:

4n + 6 = 3n + 15

4n - 3n = 15 - 6

n = 9

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If y = 2x and z = 5y, then x + 3y + z = ?

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17x
16x
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Solve for x: 2x+5=15

3
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Simplify: (3x + 2)(3x − 2)

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Find the sum: 2×1^2 +4×2^2 +6×3^ +⋯+20×10^2

76,440
77,000
77,440
6050

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Explanation

Given:

2 \times 1^^{2} + 4 \times 2^^{2} + 6 \times 3^^{2} + \dots + 20 \times 1 0^2

Write the terms as:

2 (1^^{3}) + 2 (2^^{3}) + 2 (3^^{3}) + \dots + 2 (1 0^^{3})

= 2 (1^^{3} + 2^^{3} + 3^^{3} + \dots + 1 0^^{3})

Using the formula:

1^3+2^3+cdots+n^3=left[frac{n(n+1)}{2}right]^2

For $n = 10$ :

= 2 [10 ( 11 )/2]^^{2}

= 2 (55)^^{2}

= 2 \times 3025

6050

$Last updated 27-07-2026 by NA$

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