More Math Help

[loveapples]

So I'm doing logs??in math class, and am stuck on 3 questions. >_>

1. ln 45 + ln (x-4) = ln (5x-4)

2. log (x^2 - 49) - 1 = log (x + 7)

3. log'3'(13-x) - log'3'(x+3) = 1

(That's Log base 3 (13-x) minus Log base 3 (x+3) equals to 1)

[Namine]

1. Combine the sum of ln first so:

ln [45(x-4)] = ln (5x-4)

Put everything as powers of e, so you have

e^[ln 45(x-4)] = e^[ln (5x-4)]

That'll cancel out all the ln, so you are only left with

45(x-4) = 5x-4

Just solve for x then.



2. Expand the difference of square first, so:

log [ (x+7)(x-7)] - 1 = log (x+7)

Break up the multiplication inside the log there:

log (x+7) + log (x-7) - 1 = log (x+7)

log (x-7) = 1

Since log10 = 1, x-7 must also equal to 10, so x is 17.




3. Change to base 10 (or use ln, but i hate ln) so it's easier to deal with first.

log(13-x)/log3 - log(x+3)/log3 = 1

log(13-x) - log(x+3) = log3

Again change the difference of log, combining into division for the number inside log.

log[(13-x)/(x+3)] = log3

Since both side are the same, compare the number inside the log function.

(13-x)/(x+3) = 3

And now you can solve for x.

[Zaratus]

1. ln 45 + ln (x-4) = ln (45* (x-4)), then just e^() and you have a linear equation.

2. log (x^2-49) - 1 = log (x^2-49) - log(10) = log ( (x^2-49)/10), then just 10^() and you have a quadratic equation. .. That you can solve with binomic formula like Nami there. Or do quadratic formula and get two solutions where one isn't defined for log(x+7).

Lemme think a sec about 3rd.

3. log'3'(13-x) - log'3'(x+3) = 1
(That's Log base 3 (13-x) minus Log base 3 (x+3) equals to 1)

log'3'(13-x) - log'3'(x+3) = log'3'( (13-x)/(x+3) then 3^() and linear equation again.

[loveapples]

Thanks again for the help! I appreciate it.

[mahawirasd]

wow, 2nd year of high-school seems so far away from me now... i feel old...


-w-

[Spuznik]

O.O I think my class did this today... but I was asleep for most of it, it scared me when I woke up, now I have to try to figure out what it was we were doing. :D