Solve for x: logbx = (3/2)logb4-(2/3)logb8+2logb2

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- Oct 25th 2008, 12:37 PMdc52789Solve for x
Solve for x: logbx = (3/2)logb4-(2/3)logb8+2logb2

- Oct 25th 2008, 12:40 PMChris L T521
Apply various rules of logarithms to get all the terms on the right side of the equation in

**one**logarithm.

$\displaystyle \log_bx=\log_b(8)-\log_b(4)+\log_b(4)\implies\log_bx=\log_b(?)$

I leave it for you to fill in the ?.

Then x = ?.

Can you take it from here?

--Chris

w00t!! my 12(Sun)(Sun)th post!!! (Party) - Oct 25th 2008, 01:17 PMdc52789
- Oct 25th 2008, 01:25 PMShyam
yes, x = 8

- Oct 25th 2008, 01:26 PMdc52789