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

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

Jun 2, 2015

#1
+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
+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