Answer
$x \approx 6.347$
Work Step by Step
$\log_5 x + \log_5 (2x+7) - 3 =0$
$\log_5 x + \log_5 (2x+7) =3$
$\log_5 x(2x+7) = 3$
$\log_5 (2x^{2} + 7x) = 3$
$125 = 2x^{2} + 7x$
$2x^{2} + 7x - 125 = 0$
$x = \frac{-(7)±\sqrt {7^{2}-4(2)(-125)}}{2(2)}$
$x = \frac{-7±\sqrt {49-4(2)(-125)}}{4}$
$x = \frac{-7±\sqrt {49+1000}}{4}$
$x = \frac{-7±\sqrt {1049}}{4}$
Since we can't take the log of a negative number, the only possible solution is $x = \frac{-7+\sqrt {1049}}{4}$
$x \approx 6.347$.
Check:
$\log_5 (\frac{-7+\sqrt {1049}}{4}) + \log_5 (2(\frac{-7+\sqrt {1049}}{4})+7) - 3 \overset{?}{=}0$
$\log_5 (\frac{-7+\sqrt {1049}}{4}) + \log_5 (19.6941...) - 3 \overset{?}{=}0$
$\log_5 ((\frac{-7+\sqrt {1049}}{4}) \times (19.6941...)) - 3 \overset{?}{=}0$
$\log_5 (125) - 3 \overset{?}{=}0$
$\log_5 (5^{3}) - 3 \overset{?}{=}0$
$3\log_5 (5) - 3 \overset{?}{=}0$
$3(1) - 3 \overset{?}{=}0$
$3 - 3 \overset{?}{=}0$
$0 = 0$
