SOLUTION: PLEASE HELP ME SOLVE THIS PROBLEM. Logbase3(1+x)>(1-Logbasex(1-x))Logbase3x. Thank You

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: PLEASE HELP ME SOLVE THIS PROBLEM. Logbase3(1+x)>(1-Logbasex(1-x))Logbase3x. Thank You      Log On


   



Question 24059: PLEASE HELP ME SOLVE THIS PROBLEM.
Logbase3(1+x)>(1-Logbasex(1-x))Logbase3x.
Thank You

Answer by kev82(151) About Me  (Show Source):
You can put this solution on YOUR website!

Sorry, about the slightly messy notation, but I'm still trying to figure out how to use this thing.


Ok, I assume the thing that's making it tricky for you is the rather nastly looking lg%5Bx%5D. When I first saw this problem I thought eugh! What on earth am I going to do with that. I then remembered somebody's (probably not) famous theorem.


Logs of different bases differ by a multiplicitive constant!


Let me demonstrate this with an example, say we wanted to solve 14%5Ex=29.6. I'm sure you could tell me straight away that the answer is x=lg%5B14%5D%2829.6%29. But let's try and work out the answer another way. Lets take lg%5B10%5D of both sides of the equation.


lg%5B10%5D%2814%5Ex%29=lg%5B10%5D%2829.6%29

Applying one of the many "laws of logs"


x%2Alg%5B10%5D%2814%29=lg%5B10%5D%2829.6%29

And finally dividing


x=lg%5B10%5D%2829.6%29%2Flg%5B10%5D%2814%29

So, we've solved for x in two different ways, and thus we have worked out that


lg%5B14%5D%2829.6%29=lg%5B10%5D%2829.6%29%2Flg%5B10%5D%2814%29

Now if you can do the same thing and turn the nasty lg%5Bx%5D into a much nicer lg%5B3%5D you should be able to do the problem.


There is another possibly tricky bit, but see how you get on. please write back if you didn't follow that, or you get stuck further on.


Hope that helps


Kev