**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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