To solve the equation:

$\log_2(2x + 7) = 5$

we can proceed by rewriting the equation in exponential form.

Step 1: Rewrite in Exponential Form

Since $\log_2(2x + 7) = 5$, we can rewrite this as: $2x + 7 = 2^5$

Step 2: Simplify the Exponential Expression

Calculate $2^5$: $2^5 = 32$ So now we have: $2x + 7 = 32$

Step 3: Solve for $x$

Subtract 7 from both sides: $2x = 32 - 7$ $2x = 25$

Now, divide by 2: $x = \frac{25}{2}$ $x = 12.5$

Final Answer

$x = 12.5$

Would you like further details on this solution, or have any questions?

Here are five related questions for further exploration:

1. How would you solve a logarithmic equation with a different base, like $\log_3(x + 1) = 4$?
2. What steps would you take to solve an equation involving multiple logarithmic terms, such as $\log_2(x + 3) + \log_2(x - 1) = 4$?
3. How does changing the base of the logarithm affect the solution?
4. How can you verify the solution by substituting it back into the original equation?
5. How would you solve an equation like $\log_2(2x + 7) = x$, where $x$ is also in the exponent?

Tip: Always remember to check if the solution is within the domain of the logarithmic function, as logarithms are undefined for non-positive arguments.