Math  logs
posted by
Anonymous on
.
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,
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,
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,
yep