Question

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

• A. \(2\)
• B. \(1\)
• C. \(5\)
• D. \(8\)

Hint:

Whenever a variable appears outside logarithms, convert it into logarithmic form using: \[ x=\log_b(b^x) \] This helps in combining all logarithmic expressions into a single equation.

Answer 1

The Correct Option is B

Concept:
To solve logarithmic equations, all logarithmic terms should first be expressed in the same base and then simplified using logarithmic identities. Important logarithmic identities used are: \[ a\log_b m=\log_b(m^a) \] \[ \log_b m+\log_b n=\log_b(mn) \] \[ \log_b b=1 \]

Step 1: Convert the standalone term \(x\) into logarithmic form.
We know: \[ x=\log_{15}(15^x) \] Substituting this into the given equation: \[ \log_{15}(15^x)+\log_{15}(5+3^x) = x\log_{15}5+\log_{15}24 \] Using: \[ a\log_b m=\log_b(m^a) \] we get: \[ \log_{15}(15^x)+\log_{15}(5+3^x) = \log_{15}(5^x)+\log_{15}24 \]

Step 2: Combine logarithms on both sides.
Using: \[ \log_b m+\log_b n=\log_b(mn) \] the equation becomes: \[ \log_{15}\left[15^x(5+3^x)\right] = \log_{15}(24\cdot5^x) \] Since logarithms with the same base are equal, their arguments must be equal: \[ 15^x(5+3^x)=24\cdot5^x \]

Step 3: Simplify the exponential equation.
Using: \[ 15^x=5^x\cdot3^x \] we get: \[ 5^x\cdot3^x(5+3^x)=24\cdot5^x \] Dividing both sides by \(5^x\): \[ 3^x(5+3^x)=24 \]

Step 4: Use substitution to solve the equation.
Let: \[ 3^x=t \] Then: \[ t(5+t)=24 \] \[ t^2+5t-24=0 \] Factorising: \[ (t+8)(t-3)=0 \] Hence: \[ t=-8 \quad \text{or} \quad t=3 \] But: \[ 3^x>0 \] Therefore: \[ t=-8 \] is rejected. Thus: \[ 3^x=3 \] Hence: \[ x=1 \]

Step 5: Verify the obtained solution.
Substituting \(x=1\) into the original equation: \[ 1+\log_{15}(5+3) = 1\cdot\log_{15}5+\log_{15}24 \] \[ 1+\log_{15}8 = \log_{15}5+\log_{15}24 \] Since: \[ 1=\log_{15}15 \] LHS: \[ \log_{15}15+\log_{15}8 = \log_{15}(120) \] RHS: \[ \log_{15}5+\log_{15}24 = \log_{15}(120) \] Hence both sides are equal. Therefore: \[ \boxed{x=1} \]