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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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 Ian HLv 710 months agoFavourite answer
y^2  28y + 27 = (y  1)(y 27)
3^x = 1, or, 3^x = 27
x = 0 or 3
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 ComoLv 710 months ago

Roots are found from an equation and an equation requires an equals sign and a value after the equals sign.
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 mizooLv 710 months ago
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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 10 months ago
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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