Problem 1: Solve the equation 4^2x+1=2^3x+6. & Problem 2: A bacteria culture started with a count of 480 at 8:00 A.M. and after t hours is expected to grow to f(t)=480(3/2)^t. Estimate the number of bacteria in the culture at noon the same day.
2006-11-15 18:51:22 · 12 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Ok, Victoria, if you're going to ask for help you have to write the questions correctly. Problem 1 cannot be solved using algebra the way you wrote it, I doubt you can use complex integration or any other advanced tool to solve that one. Unfortunately, I just spent the last half hour manipulating it before I remembered that little fact.
Put parentheses in. Had you put the parentheses in the problem would have been trivial. Kudos to the helpers who were smarter than I in figuring out the constants were in the exponent. It's not that hard Victoria, especially when you're asking for help.
Write the question correctly.
PS. Irma, you forgot to multiply the 2 from 4=2^2 through the whole parenthetical statement. The answer is x=4.
.
2006-11-15 19:46:44 · answer #1 · answered by ZenPenguin 7 · 2⤊ 0⤋
Problem 1 is missing parenthesis, which wouldn't be needed if the problem was written with superscript... is the problem:
4^(2x+1) = 2^(3x+6)
or
(4^2x) +1 = (2^3x) + 6
or
something else?
Since 4 = 2^2, and the law of exponents says that when raising to a power, you add exponents, you would get either:
2^(4x+1) = 2^(3x+6)
or
(2^4x) + 1 = (2^3x) + 6
I would guess that the first possibility is the most likely (and simplest answer) which would result in 4x+1 = 3x + 6, which would be x = 5
Problem 2 is much more straight forward...
The difference between 8 am and noon is four hours, so t = 4
Therefore, the function of time F(t) is equal to 480(3/2)^4
480(3/2)^4 = 480(81/16) = 2430
Just be sure to wash your hands thoroughly after handling problem number two!
2006-11-15 19:26:35 · answer #2 · answered by Irma R 2 · 0⤊ 0⤋
Problem 1:
If I'm reading it correctly, it should be 4^(2x+1)= 2^(3x+6)... so you want to use the fact that 4=2^2 and write on the left 2^(2*(2x+1)) so you will have the same base and use the fact that if a^b=a^n, then b=n. After distribution, you will have 2^(4x+2)=2^(3x+6). Set 4x+2=3x+6 and solve for x. X will equal 4
Problem 2:
The time between 8 am and noon is 4 hours. This is your time, t. Just plug 4 in for t into the equation and use your calculator to figure out 480(3/2)^4 which will end up equaling 2430.
2006-11-15 19:04:50 · answer #3 · answered by mikeyc4023 1 · 0⤊ 0⤋
are the "+1" and "+6" part of the exponents? if they are (which means it should have been 4^(2x+1) and 2^(3x+6) ) then mikey is right
1. 4^2x + 1 = 2^3x +6
(2^2)^2x +1 = 2^4x +1 = 2^3x +6
2^4x - 2^3x = 5
take the log2 on both sides
4x - 3x = x = log(base 2) 5
*note: Fiona, you made a mistake writing "log 10" for 1 and "6 log 10" for 6, you need to directly take the log of 1 and 6 (i.e. you have to take log's of all parts of the equations)
2. noon = after 4 hours
f(4) = 480 (3/2)^4
(3/2)^4 = 81/16
f(4) = 480 * (81/16) = 2430
2006-11-15 19:02:20 · answer #4 · answered by orca1006 2 · 0⤊ 0⤋
prob 1:
2x log 4 + log 10=3x log 2+6 log 10
4x log 2 - 3x log 2 = 6log10-log10
xlog2=5log10
x=5log10/log 2
x=5/log 2
prob 2:
t = 4 hours
f(4)=480(3/2)^4
f(4)=480(81/16)
f(4)=2430
hope i'm right
2006-11-15 19:06:11 · answer #5 · answered by fii 3 · 0⤊ 0⤋
you may only upload the equations at the same time and it is going to eliminate the y therefore. 4x +3y = a million +2x-3y = a million =6x=2 Divide to get x = 2/6 = a million/3. Now plug in x in between the equations to resolve for y. 4(a million/3) + 3y = a million 4/3 + 3y = a million 3y = -a million/3 y = -a million/9. If the project replaced into extra sturdy, you would could multiply between the equations through a form to get a device in which you may upload/subtract to cancel a variable.
2016-11-24 22:07:08 · answer #6 · answered by ? 4 · 0⤊ 0⤋
Have you been to Math.com??? It is this really cool site I just came across today... While studying for an entrance exam... They have everything from kindergarten math to Trig.... I'm sure you'll find what you're looking for there.... Good Luck!!!!
2006-11-15 18:55:14 · answer #7 · answered by Anonymous · 0⤊ 0⤋
Pls dont tell me you and Ohio girl are in the same algebra class? She too has been asking algebra problems!
2006-11-15 18:55:23 · answer #8 · answered by NickynJessie H 4 · 0⤊ 0⤋
4^(2x) - 2^(3x) = 5
2^(2x+1) - 2^(3x) = ln5
(2x+1)ln2 - (3x)ln2 = ln5
(2x+1-3x)ln2=ln5
(1-x)ln2 = ln5
1-x = ln5/ln2
x =1 - ln5/ln2
2006-11-15 19:37:59 · answer #9 · answered by ve1luv 2 · 0⤊ 0⤋
sorry...haviong algebraic problems myself
2006-11-15 18:53:59 · answer #10 · answered by Anonymous · 0⤊ 0⤋