|By Prethumous (Prethumous) on Thursday, June 03, 2004 - 11:49 am: Edit|
log (subscript 2) (x - 16) = log (subscript 4) (x - 4)
In the solution, one of the first steps performed transforms the above equation into the following:
log (subscript 2) (x - 16) = log (subscript 2) (x - 4) / log (subscript 2) 4
How in the world is this possible? I've probably forgotten a rule somewhere, but...
|By Optimizerdad (Optimizerdad) on Thursday, June 03, 2004 - 12:20 pm: Edit|
I'll use the Mathematically Correct :-) term 'base' instead of 'subscript' ....
There's a rule that goes log(base a) b = [log(base a) c] * [log (base c) b]
from which log(base c) b = log(base a) b / log(base a) c
Plug in a=2, b=x-4, c=4
|By Joshjmgs (Joshjmgs) on Thursday, June 03, 2004 - 12:57 pm: Edit|
log(x-4)/log(4) = log(x-16)/log(2)
[log(2)/log(4)] = log(x-16)/log(x-4)
[log(2)/log(4) = Some number A
Alog(x-4) = log(x-16)
10^A(x-4) = x-16
Sorry it was kinda long, but thats th only way I know how to use logs, im sure there is a shorter way, but, as i said, this is what I do and i have enough time with the IIc, so if youre pressed on time, you might want to make these you sides, a function and see where they intersect.
|By Benzinspeicher (Benzinspeicher) on Thursday, June 03, 2004 - 02:01 pm: Edit|
ok, i see log(sub2)(x-16)=log(sub4)(x-4)
Next, I remember my identities, and I write it out in this way: log(x-16)/log(2)=log(x-4)/log(4)
Next, I see a relationship between log(2) and log (4), which are the ones that are in the equation. The relationship I see is that log(4)=2log(2),(remember that alog(b)=log(b^a)). So, I rewrite the equation: log(x-16)/log(2)=log(x-4)/2log(2). Next, the common denominator "log(2)" cancels out. So, I'm left with log(x-16)=.5log(x-4). To make it easier for myself, I multiply both sides by 2, in order to get rid of the .5 on the right side of the equation. So, now I have 2log(x-16)=log(x-4). Next, I remember the first identity stated here (alog(b)=log(b^a)) and I see that relationship on the left side. Thus, I rewrite the equation again and get: log((x-16)^2)=log(x-4). Since these two logarithms have the same base, I can cancel out the logs. Now I am left with just plain old (x-16)^2=x-4. I use FOIL and I get x^2-32x+256=x-4. I use algebra and make the right side equal 0. I get x^2-33x+260=0. Then, I see that this is quite long, so I make this easier for myself, and I complete the square. To do this, I move the 260 over to the right. I get x^2-33x=-260. A square rootable polynomial a^2-bx+c always has c=(b/2)^2. I remember this, and I use it and get x^2-33x+16.5^2=-260 or x^2-33x+272.25=12.25 (remember, you have to add 272.25 on both sides). Now, I square root both sides, and I get x-16.5=3.5 and x-16.5=-3.5. And I solve for x, for which I get x=13 and x=20.
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