To solve question $p$:

$3^{2x} = 3^{x-1} \times 2^{4+x}$

Step-by-step solution:

1. Simplify the left-hand side (LHS) and right-hand side (RHS):

   The LHS remains $3^{2x}$. On the RHS, split it into two bases: $3^{x-1} \times 2^{4+x} = 3^{x-1} \times 2^4 \times 2^x$

   This simplifies to: $3^{x-1} \times 16 \times 2^x$

   So the equation becomes: $3^{2x} = 3^{x-1} \times 16 \times 2^x$

2. Combine powers of $3$:

   Rewrite $3^{x-1}$ as: $3^{x-1} = \frac{3^x}{3}$

   Substitute this back: $3^{2x} = \frac{3^x}{3} \times 16 \times 2^x$

   Simplify: $3^{2x} = \frac{16 \cdot 3^x \cdot 2^x}{3}$

   Multiply through by 3 to eliminate the denominator: $3 \cdot 3^{2x} = 16 \cdot 3^x \cdot 2^x$

   So: $3^{2x+1} = 16 \cdot 3^x \cdot 2^x$

3. Rewrite $3^{2x+1}$:

   Recall that $3^{2x+1} = 3^{x} \cdot 3^{x+1}$. Substituting this: $3^x \cdot 3^{x+1} = 16 \cdot 3^x \cdot 2^x$

   Cancel $3^x$ from both sides (assuming $3^x \neq 0$): $3^{x+1} = 16 \cdot 2^x$

4. Solve for $x$:

   Rewrite $3^{x+1}$ as $3 \cdot 3^x$: $3 \cdot 3^x = 16 \cdot 2^x$

   Divide both sides by $3^x$: $3 = 16 \cdot \frac{2^x}{3^x}$

   Rewrite $\frac{2^x}{3^x}$ as $\left(\frac{2}{3}\right)^x$: $3 = 16 \cdot \left(\frac{2}{3}\right)^x$

   Divide both sides by 16: $\frac{3}{16} = \left(\frac{2}{3}\right)^x$

5. Take the logarithm of both sides:

   Apply the natural logarithm (ln) to both sides: $\ln\left(\frac{3}{16}\right) = x \ln\left(\frac{2}{3}\right)$

   Solve for $x$: $x = \frac{\ln\left(\frac{3}{16}\right)}{\ln\left(\frac{2}{3}\right)}$

6. Simplify the logarithms:

   Use the logarithmic property $\ln(a/b) = \ln(a) - \ln(b)$: $x = \frac{\ln(3) - \ln(16)}{\ln(2) - \ln(3)}$

   Approximate values:

   • $\ln(3) \approx 1.0986$
   • $\ln(16) = \ln(2^4) = 4 \ln(2) \approx 4 \cdot 0.6931 = 2.7726$
   • $\ln(2) \approx 0.6931$

   Substitute: $x = \frac{1.0986 - 2.7726}{0.6931 - 1.0986}$

   Simplify the numerator and denominator: $x = \frac{-1.6740}{-0.4055}$

   Compute: $x \approx 4.13$

Final Answer:

$x \approx 4.13$

Would you like a detailed breakdown of any step or further clarification?

Related Questions:

1. How does the base logarithm choice affect solving exponential equations?
2. How do you rewrite exponential expressions like $3^{x-1}$?
3. What are logarithmic properties used in this solution?
4. Can this equation be solved graphically instead of algebraically?
5. How do we interpret $x \approx 4.13$ in a practical context?

Tip:

When solving equations involving multiple exponential terms, always try to express all terms in similar bases to simplify calculations.