Finding x with ln(a)/ln(b) as the answer. Help needed

  • Thread starter hostergaard
  • Start date
In summary: Anyway x=-3 and ln(x)=ln(2)+ln(1) In summary, the student attempted to solve a homework equation that involved the use of logarithms, but ran into trouble. They found that 3x*42x+1=4*48x and 6x+2=36x, then determined that ln(4*48x)=ln(48)*x+2*ln(2) and ln(36x)=2*ln(6)*x. However, they ran into a problem when they tried to solve for x. They determined that x=-3 and ln(x)=ln(2)+ln(1).
  • #1
hostergaard
37
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Homework Statement


http://img213.imageshack.us/img213/2327/nummer1.jpg


Homework Equations





The Attempt at a Solution


I find that that 3x*42x+1=4*48x
and 6x+2=36x

then i but ln(4*48x)=ln(48)*x+2*ln(2)
and ln(36x)=2*ln(6)*x
we get: ln(48)*x+2*ln(2)=2*ln(6)*x

Now my proplem is; Is this all correct and what how do i move them around so i get x? I I am really bad at moving those number around but x seems to disappear for me!
 
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  • #2
If you have trouble with this, simplify the equation in your head so that you have only x and constants which don't depend on x. In your case, ln(48)*x+2*ln(2)=2*ln(6)*x should look like Ax + B = Cx.
From algebra, you can add the same number or variable to both sides and preserve the equality:
Ax + B - Ax = Cx - Ax
B = Cx - Ax
When you are done, you can replace the dummies A, B and C with the real constants.
 
  • #3
then i get 2*ln(2)=2*ln(6)*x-ln(4)4*x
2*ln(2)=2*ln(3)*x


and then i move them again:
(2*ln(2))/(2*ln(3))=(ln(2))/(ln(3))=x And that should be the answer!
Thanks!
My teacher is utterly baflet over my lack of abillity to grasp some of the simpler aspect while having no trubles with some of the more advancet aspects... But that dummy tecnique i am sure will help me out with these things in the future. Thanks a lot! :-)
 
  • #4
hostergaard said:

Homework Statement


http://img213.imageshack.us/img213/2327/nummer1.jpg


Homework Equations





The Attempt at a Solution


I find that that 3x*42x+1=4*48x
and 6x+2=36x

then i but ln(4*48x)=ln(48)*x+2*ln(2)
and ln(36x)=2*ln(6)*x
we get: ln(48)*x+2*ln(2)=2*ln(6)*x

Now my proplem is; Is this all correct and what how do i move them around so i get x? I I am really bad at moving those number around but x seems to disappear for me!

[itex] 6^{x+2} = 36^x [/itex]
Let x = 3
[itex] 6^5 = 7776 [/itex]
[itex] 36^3 = 46,656 [/itex]

[itex] 6^{x+2} = 36.6^x [/itex]
 
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  • #5
.................
. 6^(x+2)=36^x - false - {a^(x+y)=(a^y)^x} .
. a^(x+y)=(a^x)*(a^y) - true
. only a^(x*y)=(a^y)*x...then 6^(x+2)=(6^x)*(6^2)=36*(6^x)
.................
.b
..y
...
...K
...a
...z
...a
...K
...
...R
...O
 
  • #6
kazak88 said:
.................
. 6^(x+2)=36^x - false - {a^(x+y)=(a^y)^x} .
. a^(x+y)=(a^x)*(a^y) - true
. only a^(x*y)=(a^y)*x...then 6^(x+2)=(6^x)*(6^2)=36*(6^x)
.................
.b
..y
...
...K
...a
...z
...a
...K
...
...R
...O

Yeah what I wrote is wrong but it was a long time ago.
 

FAQ: Finding x with ln(a)/ln(b) as the answer. Help needed

What does "Finding x with ln(a)/ln(b) as the answer" mean?

This means finding the value of x when the natural logarithm of a is divided by the natural logarithm of b.

Why would someone need to find x using ln(a)/ln(b)?

This method is used when solving exponential equations that involve variables in both the base and exponent.

What are the steps to finding x using ln(a)/ln(b)?

1. Take the natural logarithm of both sides of the equation.
2. Use the quotient property of logarithms to simplify ln(a)/ln(b) to ln(a/b).
3. Isolate the variable x by moving all other terms to the other side of the equation.
4. Use the inverse of the natural logarithm, e^x, to cancel out the ln and solve for x.

Can ln(a)/ln(b) have multiple solutions for x?

Yes, ln(a)/ln(b) can have multiple solutions for x if the equation being solved has more than one solution.

Are there any restrictions when using ln(a)/ln(b) to find x?

Yes, both a and b must be positive numbers and b cannot equal 1. Additionally, the equation being solved must have a positive solution for x.

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