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Anonymous asked in Science & MathematicsMathematics · 10 months ago

# how to answer: find the roots for f(x) when f(x) is defined as : 3^2x -28(3^x) +27?

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y^2 - 28y + 27 = (y - 1)(y -27)

3^x = 1, or, 3^x = 27

x = 0 or 3

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• -

Roots are found from an equation and an equation requires an equals sign and a value after the equals sign.

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• 3^(2x) -28(3^x) + 27

= (3^x)^2 -28(3^x) + 27

Let 3^x = a:

a^2 - 28a + 27

= (a - 1)(a -27)

a = 1, a = 27

3^x = 1 => x = 0

3^x = 27 => x = 3

The roots are x = 0, 3.

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• Let it equal 0

3^(2x) - 28 * 3^(x) + 27 = 0

Now, let 3^x = t

t^2 - 28t + 27 = 0

That's not so hard to solve, is it?

t = (28 +/- sqrt(28^2 - 4 * 27)) / 2

t = (28 +/- sqrt(2^2 * 14^2 - 4 * 27)) / 2

t = (28 +/- 2 * sqrt(196 - 27)) / 2

t = 14 +/- sqrt(169)

t = 14 +/- 13

t = 1 , 27

3^x = 1 , 27

x * ln(3) = ln(1) , ln(27)

x * ln(3) = 0 , ln(3^3)

x * ln(3) = 0 , 3 * ln(3)

x = 0/ln(3) , 3 * ln(3)/ln(3)

x = 0 , 3

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