Question

How to solve 9x-2 × 27x=320%7B%7D%5E%7Bx%7D%20%20%3D%203" id="TexFormula1" title=" { {9}^{} }^{ x - 2 } .27 {}^{x} = 3" alt=" { {9}^{} }^{ x - 2 } .27 {}^{x} = 3" align="absmiddle" class="latex-formula">


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Answer 1

Answer:

x = 1

Step-by-step 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