+ Reply to Thread
Results 1 to 7 of 7
Thread: PreCalc Help

January 29th, 2006, 12:38 PM #1
PreCalc Help
Well i need to do this online assesment of precalc question, there were about 60 questions total, and the only ones i have left are the following three. I hoping someone here can point me in the right direction because im not really sure how to do them.
Algebraic Properties of the Logarithm  Simplifying Formulas
4) Simplify ln(z^4 + z^2).
4 ln(z) + 2 ln(z) but it says this answer is incorrect
Algebraic Properties of the Logarithm  simplifying numbers
1) Simplify ln (3000).
Exponential Equations
12) Solve the following equation in x: 2 e^(x) = e^(4x  2)
e^(x)^(2) = e^(4x2) and then multiply each side by ln(x) to get
x^2 = 4x  2 > x^2  4x  2 but this doesnt appear to be the right answer so i know i went wrong somewhere.
Any help would be greatly appreciated : )

January 29th, 2006, 07:48 PM #2
someone has to know more math then i do! where are you math people?

January 29th, 2006, 09:26 PM #3
sorry. i know up to math b....

January 29th, 2006, 09:46 PM #4
I'm in PreCalc now, but we did natural logs last year and I'll be damned if I remember them
And actually, number 4 seems right to me...23. That's the number of people Mr. T has pitied in the time it has taken you to read this sentence.

January 30th, 2006, 12:39 AM #5
ln(a+b) does not equal ln(a) + ln(b) !!!
you must be thinking of ln(ab) = ln(a) + ln(b)
OK I'll take a stab....
Simplify ln(z^4 + z^2)
split into separate ln terms : 2*ln (z) + ln (z^2+1)
you can reduce further, but you need to use the imaginary number "i": 2*ln (z) + ln((z+i)(zi))
and again separate terms : 2*ln (z) + ln(z+i) + ln(zi)
Simplify ln (3000)
factor 3000 : ln (3*10^3)
more factoring: ln (3 * 5^3 * 2^3)
split into separate ln terms : ln(3) + 3*ln(5) + 3*ln(2)
not really sure if that's what you wanted though....
2 e^(x) = e^(4x  2)
now take the ln of both sides: x^2 = 4x2
rewrite into quadratic equation form: 0=x^24x+2
this polynomial is not easy to factor so you have to use the quadratic formula:
if 0 = ax^2 + bx + c) then
x = [b + sqrt(b^2  4ac)] / [2a]
pluggin in a, b, and c you get
x=[4 + sqrt(164(1)2)]/[2(1)]
x=[4 + sqrt(8)]/2
x=[4 + 2sqrt(2)]/2
x1= 2 + sqrt(2)
x2= 2  sqrt(2)
hopefully I didn't screw up somewhere...Last edited by DanU; January 30th, 2006 at 12:44 AM.

January 30th, 2006, 01:56 AM #6
Thank you so much! i am ever in your debt lol
Everything worked out perfect except for the last solution, but after doing it out i got the same exact answer as you so im not really sure whats going on with that problem. But only missing one problem out of all them isnt bad at all. So thanks again, you rock

January 30th, 2006, 05:20 AM #7
No problemo. Hope you get an "A"

Thread Information
Users Browsing this Thread
There are currently 1 users browsing this thread. (0 members and 1 guests)
Similar Threads

TI 83 plus graphing calc and Xp
By Sirnarfness in forum Applications and Operating SystemsReplies: 2Last Post: May 26th, 2004, 12:12 PM 
Calc PI
By PyroSama in forum General Tech DiscussionReplies: 4Last Post: June 12th, 2003, 02:30 AM 
Calc Program
By cracked in forum General Tech DiscussionReplies: 6Last Post: March 5th, 2003, 08:05 PM 
UD  Calc On the UD Forum
By phenious in forum Distributed ComputingReplies: 8Last Post: January 29th, 2002, 10:34 AM 
UD  Calc The First Program of 2
By phenious in forum Distributed ComputingReplies: 22Last Post: January 27th, 2002, 02:28 PM
Is It Just Me? (3350)