Posted by **Sara** on Monday, April 11, 2011 at 2:24am.

log (3x-9) = 2log3 - log27+ log(x+3)

Solving the euqation using the properties of logarithms

How do you do this?

- calculus -
**drwls**, Monday, April 11, 2011 at 5:21am
I will assume that all logs have the same base.

log(3x -9) = log [3^2*(x+3)/(3^3]

= log[(x+3)/3]

Therefore

3x -9 = (x+3)/3

9x -27 = x +3

8x = 30

x = 3.75

Check:

log (3x -9) = log 2.25 = 0.35218

2log3 -log27 + log(6.75)

= 0.95424 -1.43136 + 0.82930

= 0.35218

It makes no difference what log base you choose, as long as it is the same for all terms. (I chose base 10 when checking)

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