Question

if \\(\log_{2} (5x + 5) = 1\\), then \\(x = \\)

question 14

solve for \\(x\\):
\\(\log x + \log (x + 4) = 4\\)
\\(x = \\)

Explanation:

⚡ Using what you learned: Solving Logarithmic Equations

Step 1: Convert to exponential form

For the first equation:

$$ \log_{2}(5x + 5) = 1 $$

Convert the logarithmic equation to its exponential form:

$$ 5x + 5 = 2^1 $$

$$ 5x + 5 = 2 $$

Step 2: Solve for x

Subtract \( 5 \) from both sides:

$$ 5x = 2 - 5 $$

$$ 5x = -3 $$

Divide by \( 5 \):

$$ x = -\frac{3}{5} $$

---

Step 3: Apply product property

For the second equation (Question 14):

$$ \log x + \log(x + 4) = 4 $$

Note that \(\log\) without a base represents the common logarithm with base \( 10 \). Combine the logarithms using the product rule \(\log(a) + \log(b) = \log(ab)\):

$$ \log(x(x + 4)) = 4 $$

$$ \log(x^2 + 4x) = 4 $$

Step 4: Convert to exponential form

Convert the equation to its exponential form with base \( 10 \):

$$ x^2 + 4x = 10^4 $$

$$ x^2 + 4x = 10000 $$

Step 5: Solve the quadratic equation

Set the quadratic equation to zero:

$$ x^2 + 4x - 10000 = 0 $$

Apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

$$ x = \frac{-4 \pm \sqrt{4^2 - 4(1)(-10000)}}{2(1)} $$

$$ x = \frac{-4 \pm \sqrt{16 + 40000}}{2} $$

$$ x = \frac{-4 \pm \sqrt{40016}}{2} $$

Simplify the radical:

$$ \sqrt{40016} = \sqrt{16 \times 2501} = 4\sqrt{2501} $$

$$ x = \frac{-4 \pm 4\sqrt{2501}}{2} $$

$$ x = -2 \pm 2\sqrt{2501} $$

Step 6: Check extraneous solutions

Since the domain of \(\log x\) requires \( x > 0 \), we must discard the negative solution:

$$ x = -2 - 2\sqrt{2501} \approx -102.02 \quad \text{(extraneous)} $$

$$ x = -2 + 2\sqrt{2501} \approx 98.02 \quad \text{(valid)} $$

Answer:

For the first equation:
\( x = -\frac{3}{5} \) (or \( -0.6 \))

For Question 14:
\( x = -2 + 2\sqrt{2501} \) (or approximately \( 98.02 \))