#1**+15 **

I'm thinking you mean

$$log_3(2x-1) = 2$$

To do this, we need to rewirte this is *exponential form*!

What $$log_3(2x-1) = 2$$ is saying, is that $$3^2 = 2x-1$$

Now that we have it like this its pretty easy to solve.

3^2 = 2x-1

9 = 2x-1

10=2x

**x=5**

Here's a website explaining why you can change the log into exponets like i did. I thought they explained it well. http://www.mathsisfun.com/algebra/exponents-logarithms.html

NinjaDevo Jun 3, 2015

#1**+15 **

Best Answer

I'm thinking you mean

$$log_3(2x-1) = 2$$

To do this, we need to rewirte this is *exponential form*!

What $$log_3(2x-1) = 2$$ is saying, is that $$3^2 = 2x-1$$

Now that we have it like this its pretty easy to solve.

3^2 = 2x-1

9 = 2x-1

10=2x

**x=5**

Here's a website explaining why you can change the log into exponets like i did. I thought they explained it well. http://www.mathsisfun.com/algebra/exponents-logarithms.html

NinjaDevo Jun 3, 2015