The given equation is:

$\log 2 = \log (2x + 3)$

To solve for $x$, we can apply the property that if $\log a = \log b$, then $a = b$, as long as the bases are the same and the logs are defined. So we can set the arguments of the logs equal to each other:

$2 = 2x + 3$

Now, solve for $x$:

1. Subtract 3 from both sides: $2 - 3 = 2x$ $-1 = 2x$

2. Divide by 2: $x = -\frac{1}{2}$

So, the solution is: $x = -\frac{1}{2}$

Verification

To verify, substitute $x = -\frac{1}{2}$ back into the original expression for $2x + 3$: $2x + 3 = 2\left(-\frac{1}{2}\right) + 3 = -1 + 3 = 2$ This confirms that $\log 2 = \log (2x + 3)$ holds true.

Would you like further details or explanations?

---

Related Questions

1. How can we solve $\log a = \log b$ when additional terms are present?
2. What are the steps to simplify logarithmic equations with multiple terms?
3. How do we solve equations involving natural logarithms, $\ln$?
4. How does the base of a logarithm affect the solution process?
5. What are some applications of logarithmic equations in real-world problems?

Tip: Always check if the arguments of logarithmic expressions are positive to ensure the solution is valid.