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# 8^(1-x) = 4^(x+2)?

### 3 Answers

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• Not sure why people are disliking a seemingly correct answer.  I'll do it with an explanation and see if I get what they get:

8^(1 - x) = 4^(x + 2)

Both 4 and 8 are powers of 2:

(2³)^(1 - x) = (2²)^(x + 2)

The exponent of an exponent is the same as if you took the product of the exponents, so:

2^[3(1 - x)] = 2^[2(x + 2)]

We now have the same value on both sides with the same base on both sides so the exponents must also be the same:

3(1 - x) = 2(x + 2)

Now we can expand both sides and solve:

3 - 3x = 2x + 4

-5x = 1

x = -1/5

I do get the same answer as the others.

• 8^(1 - x) = 4^(x + 2)

[2^(3)]^(1 - x) = [2^(2)]^(x + 2) → you konw that: [x^(a)]^(b) = x^(ab)

2^[3.(1 - x)] = 2^[2.(x + 2)]

3.(1 - x) = 2.(x + 2)

3 - 3x = 2x + 4

- 5x = 1

x = - 1/5

• (2^3)^(1-x) = (2^2)^(x+2) =>

2^(3 - 3x) = 2^(2x + 4) =>

3 - 3x = 2x + 4 =>

x = -1/5.

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