Question

How to solve 9x-2 × 27x=3

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


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Answer 1

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