Solving Exponential Equations Using Logarithms & The Quadratic Formula  Summary and Q&A
TL;DR
Learn how to solve exponential equations by using logarithms and substitution.
Questions & Answers
Q: How do you simplify the equation 2^x + 4^x = 8^x?
By substituting 4 with 2^2 and 8 with 2^3, the equation becomes 2^(2x) = 2^(3x), simplifying the bases.
Q: What is the next step after simplifying the equation?
Move the terms to one side of the equation, resulting in 0 = 2^(3x)  2^(2x)  2^x.
Q: How do you solve the quadratic equation obtained from substitution?
Use the quadratic formula with a = 1, b = 1, and c = 1 to find the possible values of y.
Q: Why is only one of the values obtained from the quadratic formula valid?
An exponential equation cannot equal a negative number, so only the positive solution is valid.
Summary & Key Takeaways

The video explains how to solve an exponential equation, specifically 2^x + 4^x = 8^x.

The first step is to identify that the bases (2, 4, and 8) are all multiples of 2.

By replacing 4 and 8 with 2^2 and 2^3, the equation can be simplified to 2^(2x) = 2^(3x).

Moving the terms to one side of the equation, it becomes 0 = 2^(3x)  2^(2x)  2^x.

Dividing each term by 2^x simplifies the equation further.

By using substitution and treating y as 2^x, the quadratic equation y^2  y  1 = 0 is obtained.

Applying the quadratic formula, two possible values for y are found: 1 + √5 / 2 and 1  √5 / 2.

However, since an exponential equation cannot equal a negative number, only one solution is valid.

Using logarithms, the value of x is found to be approximately 0.694242.