Question

I'm trying to freshen up on some old math topics, but I remember very little of logarithms. Maybe you can help me out?
log(x+6) - 0.5log(2x-3) = 2 - log25
The answers are x1 = 6, x2 = 14

Answer 1

you know the basic Log rules:

1. Log A + Log B = Log (A x B)
2. Log A - Log B = Log (A/B)
3. m Log A = Log A^m
4. \[\log_{A} A = 1\]

5. Log 1 = 0

Answer 2

here in your problem the base of the log is 10. If no base is written it is taken as 10

Answer 3

@michaelbluth

Answer 4

@michaelbluth I can't sit around waiting for you.. with the rules I have written simplify your logs to one log on both sides and then you can remove the logs and just equate the quantities on both sides then solve for x

Answer 5

okay, thanks for the help, I think I know how to do this now

Answer 6

\[\log (x+6) - 0.5 \log(2x-3) = 2 - \log 25\]\[\log (x+6) - \log (2x-3)^{\frac{ 1 }{ 2 }} = 2\log 10 - \log 25\]\[\log (x +6) - \log (2x-3)^{\frac{ 1 }{ 2 }} = \log 10^{2} - \log 25\]\[\log \frac{ (x+6) }{ \sqrt{2x-3} } = \log \frac{ 100 }{ 25 }\]\[\frac{ (x+6) }{ \sqrt{2x-3} } = 4\] Now solve from here