Solve for $$ x $$ : $$ \log _{5} x+\log _{5}(x+3)=3 $$ $$ x=\square $$ Question Help: Video Submit Question

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UpStudy AI · Accepted Answer

We start with the equation:
$$
\log_{5} x + \log_{5}(x+3) = 3
$$

**Step 1:** Use the logarithm property $$\log_b a + \log_b c = \log_b (ac)$$ to combine the logarithms:
$$
\log_{5} \bigl[x(x+3)\bigr] = 3
$$

**Step 2:** Rewrite the equation in exponential form. Recall that $$\log_{b} a = c$$ means $$a = b^c$$. Thus:
$$
x(x+3) = 5^3
$$
$$
x(x+3) = 125
$$

**Step 3:** Expand and rearrange into a quadratic equation:
$$
x^2 + 3x - 125 = 0
$$

**Step 4:** Solve the quadratic equation using the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$, where $$a = 1$$, $$b = 3$$, and $$c = -125$$:
$$
x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-125)}}{2(1)}
$$
$$
x = \frac{-3 \pm \sqrt{9 + 500}}{2}
$$
$$
x = \frac{-3 \pm \sqrt{509}}{2}
$$

**Step 5:** Consider the domain restrictions for the logarithms. Since $$\log_{5} x$$ is defined only for $$x > 0$$ and $$\log_{5}(x+3)$$ for $$x+3 > 0$$, we require that:
$$
x > 0 \quad \text{and} \quad x > -3
$$
This means $$x > 0$$.

Thus, we discard the negative solution and accept:
$$
x = \frac{-3 + \sqrt{509}}{2}
$$

**Final Answer:**
$$
x = \frac{\sqrt{509}-3}{2}
$$
$$ x = \frac{\sqrt{509} - 3}{2} $$

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