# Help with logs!

1. Jan 17, 2009

### itsme03

1. How do you solve log base a ((x-1)/(x+6)) = log base a ((x-2)/(x+3))
i think you can cancel the logs and it will be ((x-1)(x+3))/((x+6)(x-2)). i don't know where to go from there.the answer is 9/2.

2. 16^x + 4^(x+1) - 3 = 0

i simplified this to:
4^(2x) + 4^(x+1) - 3 = 0

then i substituted u for 4^x and got:
u^2 + u - 3 = 0

but this cant be factored, the answers -0.315. how do i solve this?

2. Jan 18, 2009

### CompuChip

Well, "cancelling the logs" is what you are effectively doing. But if you insist on rigour, what you are doing is turning x = y into $a^x = a^y$ and then using that
$$a^{{}^a\log(x)} = x$$.
Then you get
$$\frac{x - 1}{x + 6} = \frac{x - 2}{x + 3}$$
and you can multiply crosswise (I suppose that is what you tried to do, although you wrote down something which is no longer an equation... please look up how to do this again and try to keep the equal sign in there somewhere

If you cannot factor it, then you should use the quadrature formula. However, I am wondering what happened to the " + 1" in $4^{x + 1}$.
If u = 4^x then 4^(x + 1) = ... ?

3. Jan 18, 2009

### HallsofIvy

Staff Emeritus
I disagree with CompuChip only in that "cancelling logs" is perfectly rigorous. Since logarithm is a one-to-one function log(x)= log(y) must mean x= y. However, you dropped the "= " sign! Cancelling the logs does NOT give "((x-1)(x+3))/((x+6)(x-2))", it gives (x-1)/(x+6)= (x-2)/(x+3) which is the same as (x-1)(x+3)= (x+ 6)(x- 2). Multiply that out and you get a quadratic on each side BUT the "$$x^2$$" terms cancel out leaving a simple linear equation.

No. You would get that only if u= 4^(2x) and u= 4^(x+1). You can't do that. What you can do is write 4^(2x) as (4^x)^2 and 4^(x+1)= 4(4^x) so you have (4^x)^2+ 4(4^x)- 3= 0. NOW let u= 4^x and you have u^2+ 4u- 3= 0. That can be factored.

[/quote]but this cant be factored, the answers -0.315. how do i solve this?[/QUOTE]