Posted by **Anonymous** on Saturday, January 24, 2009 at 5:49pm.

log_3(2x - 1) = 2, Find x.

Here's what I've done:

log_3(2x) * log_3(1) = 2

log2x/log3 * log1/log3 = 2

trial and error...

log2 (2*1.4)/log3 * 0 = 2.03...

x = 1.4

I think I did it wrong because anything that multiplies with 0 gives you 0...

- Math - logs -
**Damon**, Saturday, January 24, 2009 at 5:55pm
3^log_3(2x - 1) = 3^2

but 3^log3 a = a for any a

so

2 x - 1 = 3^2 = 9

2 x = 10

x = 5

- Math - logs -
**Anonymous**, Saturday, January 24, 2009 at 5:59pm
thank you! so if I get a problem like that, all I do is raise the base to the power of the log?

- Math - logs -
**Damon**, Saturday, January 24, 2009 at 6:13pm
yep

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