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# Algebra: Solving for an Immediate Value To Calculate A Desired Result

Algebra:  Solving for an Immediate Value To Calculate A Desired Result

Introduction

Considering the following problem:

We have to equations:

F = f(X)
A = a(X)

Assume that X is close to zero (0) as possible.  Therefore, we are only considering one root.

We are given the value of F and we are tasked to find A.  In order to solve for A, we must solve for F first.

Example 1:

f(X) = 2 *X
a(X) = 1/(X^3 + 3)
F = -0.04

Solving for X:
-0.04 = 2 * X
-0.02 = X

Calculating A:
A = 1/((-0.02)^3 + 3)
A = 0.3333342222

Example 2:

f(X) = e^(X^2 – 0.5)
a(X) = ln(X + 1)/(ln X + 1)
F = 10

10 = e^(X^2 – 0.5)
ln 10 = X^2 -0.5
X^2 = ln 10 + 0.5
(we’ll only use the principal root in this example)
X = √(ln 10 + 0.5)
X ≈ 1.674092319

Substituting X into a(X) to get:
A ≈ 0.6491313602

See you next time,

Eddie

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