# Math - logs

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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 -

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 -

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 -

yep

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