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Question

Find the value of $$n$$ when:
$$6^{2n+1}\div 36=6^{3}$$


Solution

$${6}^{2n + 1} \div 36 = {6}^{3}$$

$${6}^{2n + 1} \div {6}^{2} = {6}^{3}$$

$$\Rightarrow {6}^{2n + 1} \times {6}^{-2} = {6}^{3}$$

$$\Rightarrow {6}^{2n + 1 - 2} = {6}^{3}$$

By equating powers

$$\Rightarrow 2n - 1 = 3$$

$$\Rightarrow 2n = 3 + 1$$

$$\Rightarrow n = \cfrac{4}{2} = 2$$

Mathematics

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