Mathematics: Solve 81^x = 27^^^x + 2. a. x = 1 b. x = 2 c. x = 5 d. x = 6

Solve 81^x = 27^^^x + 2. a. x = 1 b. x = 2 c. x =

HernanJe6

15.03.2022

Step-by-step explanation:

81^x=27^(x+2)

3^4=81 and 3^3=27

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

exponents are the same so we just take those

4x=3x+6

x=6

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tfornwalt4390

15.03.2022

For this case we must solve the following equation:

$81 ^ x = 27^{x + 2}$

We rewrite:

$81 = 3 ^ 4\\27 = 3 ^ 3$

Then, replacing:

$3^{4x} = 3^{3 (x + 2)}$

For the bases to be equal then it must be fulfilled that:

$4x = 3 (x + 2)\\4x = 3x + 6\\4x-3x = 6\\x = 6$

x = 6

Option D

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granniejo

20.01.2022

Step-by-step explanation:

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