Solving Exponential Equations Using Logarithms & The Quadratic Formula | Summary and Q&A

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February 24, 2020

Solving Exponential Equations Using Logarithms & The Quadratic Formula

TL;DR

Learn how to solve exponential equations by using logarithms and substitution.

❓ Exponential equations can be solved by replacing bases with common exponents to simplify the equation.
🍉 Moving terms and dividing by exponents help to further simplify the equation.
❓ Substitution and the quadratic formula can be used to find possible solutions for the equation.
❓ Only positive solutions are valid for exponential equations.

Read and summarize the transcript of this video on Glasp Reader (beta).

By substituting 4 with 2^2 and 8 with 2^3, the equation becomes 2^(2x) = 2^(3x), simplifying the bases.

Move the terms to one side of the equation, resulting in 0 = 2^(3x) - 2^(2x) - 2^x.

Use the quadratic formula with a = 1, b = -1, and c = -1 to find the possible values of y.

An exponential equation cannot equal a negative number, so only the positive solution is valid.

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.