|
Requesting help from math geniuses
|
View this Thread in Original format
| cbxzcm |
Hello! I've been reviewing my work for an upcoming test, and these problems completely stump me:
1) Log(2x+5) * Log(9x^2) = 0
2) (3^x)/5^(x-1) = 2^(x-1)
3) -3+e^(x+1) = 2+e^(x-2)
4) Ln(2x-2)-Ln(x-1) = Ln(x)
5) (9^x)-7*3^x = -6
6) (Ln(8x)-2Ln(2x))/Ln(x) = 1
7) Ln(x)-Ln(x-1) = 1/2
8) e^x = 3x + 5
9) 3^x = x^3
I can't figure out how to solve for X for these problems. I have the answers to most of these problems, but I can't figure out how to get them. So, can someone please help me out? :conf: |
|
|
| lMIlk |
| number 3 is 9log900 |
|
|
| AnotherWay83 |
| quote: | Originally posted by cbxzcm
Hello! I've been reviewing my work for an upcoming test, and these problems completely stump me:
1) Log(2x+5) * Log(9x^2) = 0
|
hate to brag but here goes..
it means that either log(2x+5)=0 or log(9x^2)=0
therefore 2x+5=0 or 9x^2=0
therfore x=-5/2 or x=0
| quote: |
2) (3^x)/5^(x-1) = 2^(x-1)
|
therefore 3^x=5^(x-1)*2^(x-1)
therefore 3^x=(5*2)^(x-1)=(10)^(x-1)=10^x/10
therefore 10=10^x/3^x=(10/3)^x
take the logarithm of both sides:
log ((10/3)^x)=log 10
x*log (10/3)=1
x=1/log(10/3) :D
| quote: |
3) -3+e^(x+1) = 2+e^(x-2)
|
therefore -3 +e^x*e = 2+e^x/e^2
therefore e^x(1/e-1/e^2)=5;
therefore e^x=5*(1/e-1/e^2)
take the natural of both sides and u're dun :D
| quote: |
4) Ln(2x-2)-Ln(x-1) = Ln(x)
|
therefore Ln((2x-2)/(x-1))=Ln(x)
therefore (2x-2)/(x-1) = x
solve algebraically for x.
| quote: |
5) (9^x)-7*3^x = -6
|
this is the same as (3^2)^x - 7*3^x = -6
therefore 3^x(3^x-7) = -6
just by looking at it u can tell that x = 0.
| quote: |
6) (Ln(8x)-2Ln(2x))/Ln(x) = 1
|
therefore ln((8x/4x^2)=Ln(x)
therefore (8x/4x^2)=x
now solve for x.
| quote: |
7) Ln(x)-Ln(x-1) = 1/2
|
therefore Ln(x/(x-1))=1/2
therefore e^(1/2) = (x/(x-1))
solve for x. (e^(1/2) is just a number)
dont know about the algebraic solution but u can graph
y=Ln(3x+5)-x and see if it crosses the x-axis...that will be the solution where it crosses the x-axis..dont know if it exists tho
tell me the answer in ur textbook, and i'll tell u how to get it.
| quote: |
I can't figure out how to solve for X for these problems. I have the answers to most of these problems, but I can't figure out how to get them. So, can someone please help me out? :conf: |
hth, :D |
|
|
| Palivar |
 |
|
|
| cbxzcm |
Wow thanks for the help! But, I have a question for this solution:
| quote: | Originally posted by AnotherWay83
therefore Ln(x/(x-1))=1/2
therefore e^(1/2) = (x/(x-1))
solve for x. (e^(1/2) is just a number)
|
I got this far for this problem, but I couldn't figure out how to solve this for x algebraically. Can you explain that part? |
|
|
| AnotherWay83 |
x/(x-1)=e^(1/2)
invert both sides
(x-1)/x=e^(-1/2)---->(e^-1/2)==1/(e^(1/2))
1-1/x=e^(-1/2)
1-e^(-1/2) = 1/x
so
x=1/(1-e^(-1/2))
:D |
|
|
| cbxzcm |
| quote: | Originally posted by AnotherWay83
x/(x-1)=e^(1/2)
invert both sides
(x-1)/x=e^(-1/2)---->(e^-1/2)==1/(e^(1/2))
1-1/x=e^(-1/2)
1-e^(-1/2) = 1/x
so
x=1/(1-e^(-1/2))
:D |
How does (x-1)/x become 1-1/x ? |
|
|
| lMIlk |
| its really not that hard, you gotta pay att in class more often |
|
|
| OrZonE |
| quote: | Originally posted by AnotherWay83
hate to brag but here goes..
it means that either log(2x+5)=0 or log(9x^2)=0
therefore 2x+5=0 or 9x^2=0
therfore x=-5/2 or x=0
|
ACtually when log(x)=0 it means that x = 1 so following that logic
2x+5 = 1 or 9x^2 = 1
x=-2 or x = +-1/3
:p :p :p :p :p :thepirate |
|
|
| TBA |
e^(1/2)=x/(x-1)
e^(1/2)*x-e^(1/2)=x
(e^(1/2)*x)-x=e^(1/2)
x(e^(1/2)-1)=e^(1/2)
x=e^(1/2)/{e^(1/2)-1}
x=2.54ish |
|
|
| cbxzcm |
| quote: |
tell me the answer in ur textbook, and i'll tell u how to get it.
|
According to the book, the answer for 3^x = x^3 is 2.478 or 3. |
|
|
|
|
