Thread: 2 problems i need help on.

1. 2 problems i need help on.

couple problems i would appreciate getting help on.

9 ^ x2 = 3 ^ x/5 solve for x.

the up arrow means the following is a root.

also...

log(x^2 + 3x + 2) - 2log(x+1) ..... write as a single logarithm.

any help on either or both would be real tight.

2. Solution to log problem

log(x^2 + 3x + 2) - 2log(x+1) ..... write as a single logarithm.

Okay, so the first place to start is to rewrite the term 2log(x+1) as log(x+1)^2 because log contains the property logx^n = n*logx. So the expression now looks like:

log(x^2 + 3x + 2) - log(x+1)^2.

Another property of log is that logx - logy = log(x/y). So the above can be rewritten as:

log((x^2 + 3x +2)/(x+1)^2)

But x^2 +3 + 2 can be factored into (x+1)(x+2). So you can cancel an (x+1) term from both the numerator and denominator. The final expression should look like:

log((x+2)/(x+1)).

If I'm not mistaken I don't think it can be simplified anymore. I hope this is useful.

3. Possible solution to problem 1

9 ^ x2 = 3 ^ x/5 solve for x.

I have one question. Is 9^x2 read as 9 raised to the x squared power. Because if it is then I may have a possible solution:

9^(x^2) = 3^x/5

9^(x^2) can be rewritten as 3^(2*x^2) Since 9 = 3^2.

Now take the log of both sides:

2x^2log3 = (x/5)log3

Cancel the log 3 terms from both sides to get:

2x^2 = x/5.

subtract x/5 from both sides to get:

2x^2 - x/5 = 0

Factor out an x from each term to yield:

x(2x - 1/5) = 0.

Set each equal to zero. You get the solutions:

x = 0, 2x - 1/5 = 0
2x = 1/5
x = 1/10

x=0, x = 1/10

Perhaps you could double check with a classmate to confirm this.

4. thnx buddy