Difficulty in Exponential-solving for "X"

tekkening

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Oct 13, 2017
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Hello. I'm trying to learn actuarial analysis, but I am struggling to manipulate variables once an equation is reached. In particular, the example below: I cannot find a way to solve for "x".

((1-(1-A)^(x*B*(1-C)))*(x*(1-C)*B) * D * M * (1-F)) + G + H - I*x >= 0


I know, rationally, that it must be possible - since, through trial-and-error methodologies I can consistently find numbers that fit - however, when it comes to algebraically finding a solution, I am stymied.
 
Hello. I'm trying to learn actuarial analysis, but I am struggling to manipulate variables once an equation is reached. In particular, the example below: I cannot find a way to solve for "x".

((1-(1-A)^(x*B*(1-C)))*(x*(1-C)*B) * D * M * (1-F)) + G + H - I*x >= 0


I know, rationally, that it must be possible - since, through trial-and-error methodologies I can consistently find numbers that fit - however, when it comes to algebraically finding a solution, I am stymied.
Since you have a "power" function - you will need to use logarithm. Are you familiar with that operation? If not, start googling "power function,logarithm" and "solution of power functions" and see what did you learn. Then come back and we will see what can be done.
 
Hey guys, thank you. Let me address a few things:

Your equation seems to be overcomplicated;
for instance, (1-(1-A)) is same as (1-A).

To make a long story short:
as example, 2^x + 3x - 28 = 0 cannot be solved directly;
numeric method required.
So clearly, same applies to your equation.

And since you're solving for x, then all other variables
would be "givens", so you can greatly simplify/reduce
your equation with substitutions; as example:
K = D * M * (1-F)) + G + H

Hello! First, I would like to point out that (1-(1-A)^(x*b*(1-c)) does NOT simplify to (1-A), as the internal (1-A) has an exponent -- the exponent must resolve before we can simplify further.

(1-(1-A)^x) for example, must resolve the X before one can simply cancel the 1's out.

I am not sure I understand the rest of your post, though. You say it cannot be solved directly - does this mean "x" cannot be isolated? If it cannot, why then do specific numbers result in a solution at any given point?



Since you have a "power" function - you will need to use logarithm. Are you familiar with that operation? If not, start googling "power function,logarithm" and "solution of power functions" and see what did you learn. Then come back and we will see what can be done.

I am familiar with logarithms, but I cannot seem to reconcile a log base of 1-A with solving for X, as X is also featured in other parts of the equation.


I should clarify further - my algebraic skills are quite strong. Yet I feel as though I am missing some fundamental concept when it comes to searching for a solution, particularly in resolving equations where "x" is in both an exponent and also a product.
 
… 1 - (1 - A) is same as 1 - A …
Whoops -- there's a typo.

1 - (1 - A) = 1 + A

But, it's moot, because (as tekkening pointed out) the expression (1-A) is the base in a power, so we can't make that simplification.
 
… You say it cannot be solved directly - does this mean "x" cannot be isolated?
It means that we cannot isolate x by using the familiar algebraic manipulations.


If it cannot, why then do specific numbers result in a solution at any given point?
Oh, stating that an equation cannot be solved algebraically (i.e., directly) does not imply that no solutions exist. It simply means that we need to learn another way to find those solutions.


… I feel as though I am missing some fundamental concept when it comes to searching for a solution, particularly in resolving equations where "x" is in both an exponent and also a product.
Google keywords transcendental vs algebraic functions :cool:
 
I know, rationally, that it must be possible - since, through trial-and-error methodologies I can consistently find numbers that fit - however, when it comes to algebraically finding a solution, I am stymied.

Your x is both in an exponent and outside an exponent. No good.

Example: Solve \(\displaystyle xe^{x} = 1\)
Example: Solve \(\displaystyle e^{x} + x = 1\)

Rationality has nothing to do with it. There are numerical methods for such things. Such methods do not have to be classified as "trial and error".
 
Can you post it using Latex?
I was going to suggest that you copy and paste the left-hand side into wolfram, but wolfram could not recognize it as an expression!

Here's what MVR5 does (using lower-case symbols, as D and I are reserved for software). Click thumbnail, for larger image.

denisAid.JPG
 
It means that we cannot isolate x by using the familiar algebraic manipulations.

Oh, stating that an equation cannot be solved algebraically (i.e., directly) does not imply that no solutions exist. It simply means that we need to learn another way to find those solutions.

Google keywords transcendental vs algebraic functions :cool:
Hey guys!

I'm still struggling to isolate "x", and I've stalled at the Lambert W-Function (here: http://mathworld.wolfram.com/LambertW-Function.html).

I can't seem to find a way to account for the additional variables with it :/

Thank you again so much for the help, and my thanks continue!
 
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