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i cant figure out how to solve log_3(2x-1)=2

 Jun 2, 2015

Best Answer 

 #1
avatar+3451 
+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

 Jun 3, 2015
 #1
avatar+3451 
+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
 #2
avatar+109509 
0

Thanks for that great answer NinjaDevo  

 Jun 3, 2015

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