Please help me with logarithms?

Question

1. Rewrite 3log(x^2-4)+1/3log(x+2)-4log(x... as a single logarithm whose coefficient is 1.

2. Rewrite 2log x-3log y-5log z as a single logarithm whose coefficient is 1.

3. Solve the equation below for x.

Please note that the +3 is not in parentheses, and is therefore not part of the argument of the logarithm and 5 in log5 is the base.

log5 x+3= log5 (x-20)+4

4. Solve the equation below for x.

Please note that the +5 is not in parentheses, and is therefore not part of the argument of the logarithm and 2 in log2 is the base.

log2 x+5=8-log2(x+7)

5. Solve the equation below for x. The x in logx is the base.
logx 9= -2

I really need some help with these cause i have never in my life used this before. i appreciate it guys. thanks

Answer 1

1) = log[((x^2-4)^3 *(x+2)^1/3)/x^4]
2)follow example 1)
3) log[x/(x-20)]=1 so by definition x/(x-20) =5 ( your base)
x=5x-100 4x=100 and x = 25
Now I think you can carry on

Answer 2

hint: opposite powers via using logarithms. So opposite a^n via taking the log of a^n with a base a; n=log[a](a^n) And opposite log base a via taking the potential of a. a^(log[a](n))=n.

Answer 3

Do your own homework. I got myself through calculus, not about to carry your weight too. Get a tuitor or send me money.