Solving Logarithmic Equations

Today we'll be solving logarithmic equations. To begin, who can tell me the relationship between logarithms and exponents?

Logarithms are the opposite of exponents!

Correct! If we have $log_a(b) = c$, it means $a^c = b$. This is important for solving equations. What would be our first step to solve $log_x = 2$?

We can convert it to exponential form, so it becomes $x = 10^2$.

Exactly! This gives us the solution $x = 100$. Remember, converting to exponential form is key in solving these problems.

Let's try an equation: $log(x+2) = 1$. Who wants to solve it?

We convert it to exponential, so it's $x + 2 = 10^1$, which means $x + 2 = 10$.

Great! What do we do next?

We subtract 2 from both sides to find $x = 8$.

Perfect! This application of logarithms allows us to solve equations step by step.

Now, let's address a more complex example: $log(x) + log(x-3) = 1$. How can we simplify this?

We can use the product rule to combine the logs into one: $log[x(x-3)] = 1$.

Exactly! What does this conversion tell us?

It means $x(x-3) = 10^1$ or $x^2 - 3x - 10 = 0$.

Bingo! Now we solve the quadratic. Remember to check for valid solutions.

Lastly, after solving, why do we need to check our solutions?

To make sure that the values we find make the logarithms valid!

Exactly! Since logarithms must have positive arguments, any invalid solutions will be excluded from our final answers.

So, we only keep valid answers!

Correct! Remember, valid arguments are key to logarithmic equations.