Logs

[Zfuss12]

ive been having trouble with this problem and was wondering if I could get some help, it has been awhile since I have tackled logs:

2logbase3x-logbase3(x+1)=1

 

[tkhunny]

ive been having trouble with this problem and was wondering if I could get some help, it has been awhile since I have tackled logs:

2logbase3x-logbase3(x+1)=1

Use the rules you have.

Start with this one: \(\displaystyle a*log(b) = log(b^{a})\)

Then use this one. \(\displaystyle log(b) - log(c) = log(\frac{b}{c})\)

Finish up with this one: \(\displaystyle log_b(a) = c \Rightarrow b^{c} = a\)

After that, it's a tiny algebra problem.

 

[steve_b]

Just to follow on to tk's answer, after you finish those steps, you will get a quadratic equation:

x^2 = 3x + 3

or,

x^2 - 3x - 3 = 0

Solving this (by your favorite method), you get:

x = [3+sqrt(21)] / 2 OR x = [3-sqrt(21)] / 2

The second of these solutions has to be discarded because when you substitute it into the original equation, you try to take the log of a negative number, which (withour complex numbers) you can't do.

Steve

 

[Zfuss12]

Thanks alot for everyones help, through your guidance and help I can use this problem as a template for it shows me many of the rules and finish the rest of the problems......thanks for the kick start