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Solving Natural Exponential Functions 3 Examples with Natural Logarithms
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In this video I solve 3 equations that involve base e exponential functions using natural logarithms. Example 1 at 1:02 Example 2 solved with 2 methods are at 6:18 and 11:24 Example 3 solved two ways at 13:56 and 17:10
I got some more great suggestions for solving my examples from Jay Ballauer,
Second Example:
Rewrite e^2x at power-of-a-power (e^x)^2 first...then you have a pair of e^x, which you let u = e^x. That let's students see the quadratic because they see e^x as replaceable (there are two of them, which is easy to see). Once you work u^2 -6u-7 = 0 as a quadratic, you can then replace u with e^x at the end of the problem for both solutions.
It's much cleaner that way and its a helpful method for many types of problems...like when they see the same thing with trig functions in a couple of months. Of course you don't have to use "u" as a variable, but its traditional for substitution.
Third Example:
On the third example, you can just divide both sides by e^(1-x) from the beginning. For the left side you get e^(3x+1)/e^(1-x) which is simple division of same bases...so you subtract powers to get e^(3x+1-(1-x)), or just e^4x (watch out for the distribution of negative there). So, you get to e^4x= 6 after a couple of easy steps.
I got some more great suggestions for solving my examples from Jay Ballauer,
Second Example:
Rewrite e^2x at power-of-a-power (e^x)^2 first...then you have a pair of e^x, which you let u = e^x. That let's students see the quadratic because they see e^x as replaceable (there are two of them, which is easy to see). Once you work u^2 -6u-7 = 0 as a quadratic, you can then replace u with e^x at the end of the problem for both solutions.
It's much cleaner that way and its a helpful method for many types of problems...like when they see the same thing with trig functions in a couple of months. Of course you don't have to use "u" as a variable, but its traditional for substitution.
Third Example:
On the third example, you can just divide both sides by e^(1-x) from the beginning. For the left side you get e^(3x+1)/e^(1-x) which is simple division of same bases...so you subtract powers to get e^(3x+1-(1-x)), or just e^4x (watch out for the distribution of negative there). So, you get to e^4x= 6 after a couple of easy steps.
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