Exponential Equations - Algebra and Precalculus

This Algebra and Precalculus video tutorial explains how to solve difficult exponential equations.

Log to Exponential Form:

Change of Base Formula:

Change of Base Log Problem:

Properties of Logarithms:

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Expanding Logarithmic Expressions:

Condensing Logarithmic Expressions:

Natural Logarithms:

Solving Exponential Equations:

Exponential Equations - Quadratic Form:

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Solving Logarithmic Equations:

Graphing Logarithmic Functions:

Graphing Exponential Functions:

Compound Interest Word Problems:

Logarithms Practice Problems:

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Final Exams and Video Playlists:

Full-Length Videos and Worksheets:

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[OwaisAkber]

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[siddyboi]

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[superb]

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[georgesadler]

MR. Organic Chemistry Tutor, thank you for explaining and solving Exponential Equations.

[afikanto]

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[benign]

Am that noob that'd use logarithms but nice video on the alternative method.

[oZqdiac]

For the first one
I split 3^(x + 2) into 3^x * 3^2 like you did to get (3^x * 3^2) + 3^x = 10/81
Then I let u = 3^x
Which now becomes (u * 3^2) + u = 10/81
3^2 = 9
So it is now (u * 9) + u = 10/81
Which is written as 9u + u = 10/81
Which can be written as 10u = 10/81
Divide both sides by 10 to get u = 1/81
Remember u = 3^x so 3^x = 1/81
81 can be written as 3^4 so you have 3^x = 1/(3^4)
Aha! This is the reciprocal of 3^4! Which means we can write this as 3^-4 as a negative power means the reciprocal of the same number with the positive power.
So 3^x = 3^-4
Bases are same so powers must be same
x = -4

Proof:
3^(-4 + 2) + 3^(-4) = 10/81
3^(-2) + 3^(-4) = 10/81
1/9 + 1/81 = 10/81
1/9 = 9/81
9/81 + 1/81 = 10/81
10/81 = 10/81
OK