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How to solve 9x-2 × 27x=3

{ {9}^{} }^( x - 2 ) .27 {}^(x) = 3




User Jagira
by
7.7k points

1 Answer

7 votes

Answer:

x = 1

Explanation:

Using the rules of exponents


(a^m)^(n) =
a^(mn)


a^(m) ×
a^(n)
a^((m+n))

Note that 9 = 3² and 27 = 3³ , thus


9^(x-2) ×
27^(x)

=
(3^2)^(x-2) ×
(3^3)^(x)

=
3^(2x-4) ×
3^(3x)

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

=
3^(5x-4) , then


3^(5x-4) =
3^(1)

Since the bases on both sides are equal, equate the exponents

5x - 4 = 1 ( add 4 to both sides )

5x = 5 ( divide both sides by 5 )

x = 1

User Algot
by
9.0k points

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