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

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Homework Help Overview

The discussion revolves around solving an equation involving logarithms, specifically finding the value of x in the context of exponential expressions. The original poster presents an equation derived from manipulating exponential forms and logarithmic identities.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the manipulation of logarithmic equations and question the correctness of the original poster's steps. There are attempts to simplify the equation and isolate x, with some participants suggesting algebraic techniques for rearranging terms.

Discussion Status

The discussion includes various attempts to clarify the steps needed to isolate x, with some participants providing algebraic insights. There is acknowledgment of confusion regarding the manipulation of logarithmic expressions, and the original poster expresses gratitude for the guidance received.

Contextual Notes

Some participants note the original poster's struggle with basic algebraic manipulation despite understanding more advanced concepts. There are references to specific equations and identities that may not have been applied correctly, leading to further questions about assumptions in the problem setup.

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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!
 
Last edited by a moderator:
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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.
 
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! :-)
 
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]
 
Last edited by a moderator:
.................
. 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
 
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.
 

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