If $(2^{3 x - 1} + 10) \div 7 = 6$ , then $x$ is equal to

(a) $- 2$

(b) $0$

(c) $1$

(d) $2$

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Solution

Correct option is (d)
Given: $[2^{3 x - 1} + 10] \div 7 = 6$
$\Rightarrow 2^{3 x - 1} + 10 = 6 \times 7$
$\Rightarrow 2^{3 x - 1} + 10 = 42$

$\Rightarrow 2^{3 x - 1} = 42 - 10$
$\Rightarrow 2^{3 x - 1} = 32$
$\Rightarrow 2^{3 x - 1} = (2)^{5}$
When bases are same across equal sign then powers must be also same.
$\Rightarrow 3 x - 1 = 5$
$\Rightarrow 3 x = 6$

$∴ x = \frac{6}{3} = 2$

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