Question

Solve the equation: \[ x + \log_{15}(5 + 3x) = x \log_{15} 5 + \log_{15} 24 \]

Hint:

When dealing with logarithmic equations, always simplify using log properties such as: \[ \log_b A + \log_b B = \log_b(AB), \quad \log_b A^n = n \log_b A \] Try substitution if algebraic manipulation gets messy.

Answer 1

We are given the equation: \[ x + \log_{15}(5 + 3x) = x \log_{15} 5 + \log_{15} 24 \] Step 1: Bring like terms together Move all terms to one side: \[ x - x \log_{15} 5 + \log_{15}(5 + 3x) - \log_{15} 24 = 0 \] Step 2: Use log rules Use the subtraction property: \[ \log_b A - \log_b B = \log_b \left(\frac{A}{B}\right) \] So: \[ x(1 - \log_{15} 5) + \log_{15} \left( \frac{5 + 3x}{24} \right) = 0 \] Now move one term to the other side: \[ \log_{15} \left( \frac{5 + 3x}{24} \right) = -x(1 - \log_{15} 5) \] Let’s try plugging in the options. --- Try \( x = 2 \): Left-hand side: \[ \log_{15}(5 + 3 \cdot 2) = \log_{15}(11) \] Right-hand side: \[ 2 \log_{15} 5 + \log_{15} 24 \] Left: \[ 2 + \log_{15} 11 \] Right: \[ 2 \log_{15} 5 + \log_{15} 24 \] Now test LHS: \[ 2 + \log_{15} 11 \] RHS: \[ = \log_{15} 5^2 + \log_{15} 24 = \log_{15}(25 \cdot 24) = \log_{15}(600) \] Now LHS: \[ = \log_{15}(15^2) + \log_{15} 11 = \log_{15}(225 \cdot 11) = \log_{15}(2475) \] But this gets messy. Try solving algebraically instead. --- Try direct substitution of options into original equation. Option A: \( x = 2 \) LHS: \[ x + \log_{15}(5 + 3x) = 2 + \log_{15}(5 + 6) = 2 + \log_{15}(11) \] RHS: \[ x \log_{15} 5 + \log_{15} 24 = 2 \log_{15} 5 + \log_{15} 24 = \log_{15}(25) + \log_{15}(24) = \log_{15}(600) \] Now check if: \[ 2 + \log_{15}(11) = \log_{15}(600) \Rightarrow \log_{15}(15^2 \cdot 11) = \log_{15}(2475) \] So both sides: \[ \log_{15}(2475) = \log_{15}(600) \Rightarrow \text{False} \] Wait — this contradiction implies \( x = 2 \) is NOT the solution. Let’s instead plug values into the full original expression: Try Option B: \( x = 1 \) LHS: \[ 1 + \log_{15}(5 + 3 \cdot 1) = 1 + \log_{15}(8) \] RHS: \[ 1 \cdot \log_{15} 5 + \log_{15} 24 = \log_{15}(5 \cdot 24) = \log_{15}(120) \] LHS: \[ = \log_{15}(15) + \log_{15}(8) = \log_{15}(15 \cdot 8) = \log_{15}(120) \] Both sides equal → ✅ \[ \boxed{x = 1} \]