## Intermediate Algebra: Connecting Concepts through Application

$x \approx 2.961$
$\log_2 (2x) + \log_2 (5x-4) = 6$ $\log_2 2x(5x-4) = 6$ $2^{6} = 2x(5x -4)$ $64 = 2x(5x -4)$ $64 = 10x^{2} - 8x$ $10x^{2} - 8x - 64 = 0$ $2(5x^{2} - 4x - 32) = 0$ $x = \frac{-(-4)±\sqrt {(-4)^{2}-4(5)(-32)}}{2(5)}$ $x = \frac{4±\sqrt {16-4(5)(-32)}}{10}$ $x = \frac{4±\sqrt {16+640}}{10}$ $x = \frac{4±\sqrt {656}}{10}$ $x = \frac{4±4\sqrt {41}}{10}$ $x = \frac{2±2\sqrt {41}}{5}$ $x \approx 2.961, -2.161$ $x \approx 2.961$ Since $x$ can't be negative, the only possible solution is $x \approx 2.961$. This is because $\log$ negative a number does not exist and therefore $x$ can only be a positive number in this case. Check: $\log_2 (2( \frac{2+2\sqrt {41}}{5})) + \log_2 (5(\frac{2+2\sqrt {41}}{5})-4) \overset{?}{=} 6$ $\log_2 (5.9224...) + \log_2 (10.8062...) \overset{?}{=} 6$ $\log_2 64 \overset{?}{=} 6$ $\log_2 2^{6} \overset{?}{=} 6$ $6\log_2 2 \overset{?}{=} 6$ $6(1) \overset{?}{=} 6$ $6 = 6$