Let’s solve these equations step by step:

a) $\frac{2 \log(x+1)}{\log 3} = 2x + 1$

Step 1: Isolate the logarithmic term $2 \log(x+1) = (2x + 1)\log 3$ Divide through by 2: $\log(x+1) = \frac{(2x + 1)\log 3}{2}$

Step 2: Rewrite the logarithmic equation in exponential form $x+1 = 10^{\frac{(2x + 1)\log 3}{2}}$

Simplify $10^{\log 3} = 3$: $x+1 = 3^{\frac{2x+1}{2}}$

Step 3: Solve for $x$ Now, we need to solve $x + 1 = 3^{\frac{2x+1}{2}}$. This can be solved either graphically or numerically, as it does not simplify further algebraically.

b) $8^{x+12} = 2^{x^2 + 3x}$

Step 1: Rewrite bases as powers of 2 $8^{x+12} = (2^3)^{x+12} = 2^{3(x+12)}$ $2^{3(x+12)} = 2^{x^2 + 3x}$

Step 2: Equate exponents Since the bases are the same, equate the exponents: $3(x+12) = x^2 + 3x$ Expand and simplify: $3x + 36 = x^2 + 3x$ $x^2 - 36 = 0$

Step 3: Solve the quadratic equation Factorize: $(x - 6)(x + 6) = 0$ Thus: $x = 6 \quad \text{or} \quad x = -6$

Step 4: Verify solutions For logarithmic or exponential equations, verify solutions in the original equation:

• $x = -6$ is not valid since it leads to negative exponents in the original equation.
• Therefore, the only solution is: $x = 6$

Would you like me to further analyze the solutions, explain numerical methods for (a), or explore alternative approaches?

Related Questions:

1. How can we use logarithmic identities to simplify equations involving $\log(x)$?
2. What numerical methods can solve $x+1 = 3^{\frac{2x+1}{2}}$?
3. How do we determine the validity of solutions in exponential equations?
4. Can this approach be generalized to other bases in exponential equations?
5. What are the common pitfalls when solving equations with mixed exponents and logarithms?

Tip:

Always verify your solutions in the original equations, especially in logarithmic and exponential equations, to avoid introducing extraneous solutions.