Question 177(v)

Find x.
$2^{3 x} = 8^{2 x + 1}$

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Solution

We have,
$2^{3 x} = 8^{2 x + 1} \Rightarrow 2^{3 x} = (2^{3})^{2 x + 1} \Rightarrow 2^{3 x} = (2)^{6 x + 3}$ [ $∵ (a^{m})^{n} = (a)^{m \times n}$ ]

Since the base is same on both sides, we can equate the powers.
On equating the powers of 2, we get
3x = 6x + 3
$\Rightarrow$ 3x - 6x= 3
$\Rightarrow$ -3x = 3
$\Rightarrow x = 3 \div (- 3)$
$\Rightarrow x = - 1$

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Question 177(v)

Find x.
$2^{3 x} = 8^{2 x + 1}$

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