Here's the question you clicked on:
sofia_smr
Solve 3^(x – 4) = 7^(x + 9)
what to do wuth this question? Which Chapter Question??
i am in florida virtualll....
name the chapter or topic of your book from which this question u have asked!!
OK, so you're having trouble with the same stuff. 1) please don't keep repeat-posting your questions. That's against our Code of Conduct 2) What specifically are you having trouble with. The entire theory?
sorry laura. Where can i check the code of cunduct so i dont make that mistake?
ok thank you . And are you like in charge of the program?
I'm one of the staff members here on site, yes.
ohh. And you asked me what i was having trouble with... mostly with the entriel theory of logs actuall :S
do you think you can help me with logs?
No, I really can't help with logs. Sorry. I don't remember most of this math....
ohhh, do who know any group member that can help me with this?
@sofia I'd help you. Do you need an explanation on logarithms?
@sofia_smr What help you want with logs?
Take LN on both sides: Ln(3^(x-4))=Ln(7^(x+9)) Using the property Lna^b=b*Lna: (x-4)Ln3=(x+9)Ln7
Input Interpretation: solve ; 3 ^ x-4 = 7 ^ x+9 Results: this soo hard to type because of lag and error equation so i post a site that just show you the answer and hopefully you know how to do the equation http://www.wolframalpha.com/bing/?i=solve+3%5e(x+-+4)+%3d+7%5e(x+%2b+9)&expl=3
i cant understand pretty much everything about them :S
it shows results, real solution and plot.
Distribute and simplify.. i gotta go to class bbl
3^(x-4)=7^(x+9) Take the natural logarithm of both sides of the equation to remove the variable from the exponent. ln(3^(x-4))=ln(7^(x+9)) The left-hand side of the equation is equal to the exponent of the logarithm argument because the base of the logarithm equals the base of the argument. xln(3)-4ln(3)=ln(7^(x+9)) The exponent of a factor inside a logarithm can be expanded to the front of the expression using the third law of logarithms. The third law of logarithms states that the logarithm of a power of x is equal to the exponent of that power times the logarithm of x (e.g. log^b(x^(n))=nlog^b(x)). xln(3)-4ln(3)=((x+9)ln(7)) Multiply ln(7) by each term inside the parentheses. xln(3)-4ln(3)=(xln(7)+9ln(7)) Since xln(7) contains the variable to solve for, move it to the left-hand side of the equation by subtracting xln(7) from both sides. xln(3)-4ln(3)-xln(7)=9ln(7) Factor out the GCF of x from each term in the polynomial. x(ln(3))+x(-ln(7))=4ln(3)+9ln(7) Factor out the GCF of x from xln(3)-xln(7). x(ln(3)-ln(7))=4ln(3)+9ln(7) Divide each term in the equation by (ln(3)-ln(7)). (x(ln(3)-ln(7)))/(ln(3)-ln(7))=(4ln(3))/(ln(3)-ln(7))+(9ln(7))/(ln(3)-ln(7)) Simplify the left-hand side of the equation by canceling the common factors. x=(4ln(3))/(ln(3)-ln(7))+(9ln(7))/(ln(3)-ln(7)) Simplify the right-hand side of the equation by simplifying each term. x=(4ln(3)+9ln(7))/(ln(3)-ln(7))
ok thanks a lot though, good luck
(x-4)log3 = (x+9)log7 Sofia, sorry i don't gotta calculator :/
@sofia_smr Did you understand?
yes! shaayaan gave me a great example there!
glad you got help by someone :)