+3454 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
+3454 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
