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Chapter 13: Q. 8 (page 870)

Solve: $\log_{2} (3 x - 2) + \log_{2} x = 4$

Short Answer

Expert verified

The solution of the given equation: $x = - 1, \frac{8}{3}$

Step by step solution

01

Step 1. Given information

Given equation:

$\log_{2} (3 x - 2) + \log_{2} x = 4$

02

Step 2. Solve the equation.

$\log_{2} (3 x - 2) + \log_{2} x = 4 \log_{2} (3 x - 2) (x) = 4$

Take antilog on both sides.

$x (3 x - 2) = 2^{4} 3 x^{2} - 2 x = 16 3 x^{2} - 2 x - 16 = 0 3 x^{2} - 8 x + 6 x - 16 = 0 x (3 x - 8) + 2 (3 x - 8) = 0 (3 x - 8) (x + 2) = 0 3 x - 8 = 0 x = \frac{8}{3} x + 2 = 0 x = - 2$

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