Elbert is studying properties of exponents in his algebra class. His teacher challenged the class to write an algebraic equation that used exponent properties and had at least four different answers.

Elbert did even better - he used absolute value and two basic exponent properties that involve equaling 1 to make an equation that had six different answers! Can you solve Elbert's equation and find all six answers?

Solve __2__ ways... __1 algebraic____ __and __1 non algebraic__

http://mrperalta.com/wp-content/uploads/2019/04/Q4-Project-PoW-3-Elberts-Equation.pdf

PS - I didnt know how to add the equation. So there is the link to the question.

Guest Apr 19, 2019

#1**0 **

Solve for x over the real numbers:

abs(2 x^2 + 3 x - 1)^(3 x^2 - 13 x + 4) = 1

Take the natural logarithm of both sides:

log(abs(2 x^2 + 3 x - 1)) (3 x^2 - 13 x + 4) = 0

The left hand side factors into a product with three terms:

log(abs(2 x^2 + 3 x - 1)) (x - 4) (3 x - 1) = 0

Split (x - 4) (3 x - 1) log(abs(2 x^2 + 3 x - 1)) into separate parts with additional assumptions.

Assume abs(2 x^2 + 3 x - 1)!=0 from log(abs(2 x^2 + 3 x - 1)):

x - 4 = 0 for abs(2 x^2 + 3 x - 1)!=0

or 3 x - 1 = 0 for abs(2 x^2 + 3 x - 1)!=0

or log(abs(2 x^2 + 3 x - 1)) = 0

Add 4 to both sides:

x = 4 or 3 x - 1 = 0 for abs(2 x^2 + 3 x - 1)!=0

or log(abs(2 x^2 + 3 x - 1)) = 0

Add 1 to both sides:

x = 4 or 3 x = 1 for abs(2 x^2 + 3 x - 1)!=0

or log(abs(2 x^2 + 3 x - 1)) = 0

Divide both sides by 3:

x = 4 or x = 1/3 or log(abs(2 x^2 + 3 x - 1)) = 0

Cancel logarithms by taking exp of both sides:

x = 4 or x = 1/3 or abs(2 x^2 + 3 x - 1) = 1

Split the equation into two possible cases:

x = 4 or x = 1/3 or 2 x^2 + 3 x - 1 = 1 or 2 x^2 + 3 x - 1 = -1

Divide both sides by 2:

x = 4 or x = 1/3 or x^2 + (3 x)/2 - 1/2 = 1/2 or 2 x^2 + 3 x - 1 = -1

Add 1/2 to both sides:

x = 4 or x = 1/3 or x^2 + (3 x)/2 = 1 or 2 x^2 + 3 x - 1 = -1

Add 9/16 to both sides:

x = 4 or x = 1/3 or x^2 + (3 x)/2 + 9/16 = 25/16 or 2 x^2 + 3 x - 1 = -1

Write the left hand side as a square:

x = 4 or x = 1/3 or (x + 3/4)^2 = 25/16 or 2 x^2 + 3 x - 1 = -1

Take the square root of both sides:

x = 4 or x = 1/3 or x + 3/4 = 5/4 or x + 3/4 = -5/4 or 2 x^2 + 3 x - 1 = -1

Subtract 3/4 from both sides:

x = 4 or x = 1/3 or x = 1/2 or x + 3/4 = -5/4 or 2 x^2 + 3 x - 1 = -1

Subtract 3/4 from both sides:

x = 4 or x = 1/3 or x = 1/2 or x = -2 or 2 x^2 + 3 x - 1 = -1

Divide both sides by 2:

x = 4 or x = 1/3 or x = 1/2 or x = -2 or x^2 + (3 x)/2 - 1/2 = -1/2

Add 1/2 to both sides:

x = 4 or x = 1/3 or x = 1/2 or x = -2 or x^2 + (3 x)/2 = 0

Factor x and constant terms from the left hand side:

x = 4 or x = 1/3 or x = 1/2 or x = -2 or 1/2 x (2 x + 3) = 0

Multiply both sides by 2:

x = 4 or x = 1/3 or x = 1/2 or x = -2 or x (2 x + 3) = 0

Split into two equations:

x = 4 or x = 1/3 or x = 1/2 or x = -2 or x = 0 or 2 x + 3 = 0

Subtract 3 from both sides:

x = 4 or x = 1/3 or x = 1/2 or x = -2 or x = 0 or 2 x = -3

Divide both sides by 2:

**x = 4 or x = 1/3 or x = 1/2 or x = -2 or x = 0 or x = -3/2**

Guest Apr 19, 2019

#1**0 **

Best Answer

Solve for x over the real numbers:

abs(2 x^2 + 3 x - 1)^(3 x^2 - 13 x + 4) = 1

Take the natural logarithm of both sides:

log(abs(2 x^2 + 3 x - 1)) (3 x^2 - 13 x + 4) = 0

The left hand side factors into a product with three terms:

log(abs(2 x^2 + 3 x - 1)) (x - 4) (3 x - 1) = 0

Split (x - 4) (3 x - 1) log(abs(2 x^2 + 3 x - 1)) into separate parts with additional assumptions.

Assume abs(2 x^2 + 3 x - 1)!=0 from log(abs(2 x^2 + 3 x - 1)):

x - 4 = 0 for abs(2 x^2 + 3 x - 1)!=0

or 3 x - 1 = 0 for abs(2 x^2 + 3 x - 1)!=0

or log(abs(2 x^2 + 3 x - 1)) = 0

Add 4 to both sides:

x = 4 or 3 x - 1 = 0 for abs(2 x^2 + 3 x - 1)!=0

or log(abs(2 x^2 + 3 x - 1)) = 0

Add 1 to both sides:

x = 4 or 3 x = 1 for abs(2 x^2 + 3 x - 1)!=0

or log(abs(2 x^2 + 3 x - 1)) = 0

Divide both sides by 3:

x = 4 or x = 1/3 or log(abs(2 x^2 + 3 x - 1)) = 0

Cancel logarithms by taking exp of both sides:

x = 4 or x = 1/3 or abs(2 x^2 + 3 x - 1) = 1

Split the equation into two possible cases:

x = 4 or x = 1/3 or 2 x^2 + 3 x - 1 = 1 or 2 x^2 + 3 x - 1 = -1

Divide both sides by 2:

x = 4 or x = 1/3 or x^2 + (3 x)/2 - 1/2 = 1/2 or 2 x^2 + 3 x - 1 = -1

Add 1/2 to both sides:

x = 4 or x = 1/3 or x^2 + (3 x)/2 = 1 or 2 x^2 + 3 x - 1 = -1

Add 9/16 to both sides:

x = 4 or x = 1/3 or x^2 + (3 x)/2 + 9/16 = 25/16 or 2 x^2 + 3 x - 1 = -1

Write the left hand side as a square:

x = 4 or x = 1/3 or (x + 3/4)^2 = 25/16 or 2 x^2 + 3 x - 1 = -1

Take the square root of both sides:

x = 4 or x = 1/3 or x + 3/4 = 5/4 or x + 3/4 = -5/4 or 2 x^2 + 3 x - 1 = -1

Subtract 3/4 from both sides:

x = 4 or x = 1/3 or x = 1/2 or x + 3/4 = -5/4 or 2 x^2 + 3 x - 1 = -1

Subtract 3/4 from both sides:

x = 4 or x = 1/3 or x = 1/2 or x = -2 or 2 x^2 + 3 x - 1 = -1

Divide both sides by 2:

x = 4 or x = 1/3 or x = 1/2 or x = -2 or x^2 + (3 x)/2 - 1/2 = -1/2

Add 1/2 to both sides:

x = 4 or x = 1/3 or x = 1/2 or x = -2 or x^2 + (3 x)/2 = 0

Factor x and constant terms from the left hand side:

x = 4 or x = 1/3 or x = 1/2 or x = -2 or 1/2 x (2 x + 3) = 0

Multiply both sides by 2:

x = 4 or x = 1/3 or x = 1/2 or x = -2 or x (2 x + 3) = 0

Split into two equations:

x = 4 or x = 1/3 or x = 1/2 or x = -2 or x = 0 or 2 x + 3 = 0

Subtract 3 from both sides:

x = 4 or x = 1/3 or x = 1/2 or x = -2 or x = 0 or 2 x = -3

Divide both sides by 2:

**x = 4 or x = 1/3 or x = 1/2 or x = -2 or x = 0 or x = -3/2**

Guest Apr 19, 2019