Confused with logarithms?

Question

Confused with logarithms?

i havent done logarithms in a while and we're reviewing it right now so im having some trouble.can some1 help me out with this question

I have to solve for x:

logx+log(x-3)=1

and

2log(x-3)+log(x+2)-6logx

and please give me an explanation of how to solve them.thanks

Answer 1

First use the rules of logarithms to combine the equation into one log:

log x*(x-3)=1

Now, this means 10^[x*(x-3)]=1

Since 10^0=1, x*(x-3) must equal 0

x*(x-3)=0

x=0 or x=3

Rules of logarithms:

2log a = log (a^2)

log a + log b = log a*b

log a - log b = log a/b

use these rules to combine the three logs into one expression.

log[(x-3)^2]+log(x+2) - log (x^6)

log [(x+2)(x+3)^2/x^6]

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