Answer:
x = [tex]$ \frac{-5}{2} $[/tex]
Step-by-step explanation:
Given: [tex]$ 3^{4x - 5} = (\frac{1}{27})^{2x + 10} $[/tex]
Since, xᵃ . xᵇ = xᵃ ⁺ ᵇ we have
[tex]$ 3^{4x} . 3^{-5} $[/tex] = [tex]$ \frac{1}{27^{2x + 10}} $[/tex]
Since, 3³ = 27, we have: [tex]$ 3^{4x} . 3^{-5} = \frac{1}{3^{(3)(2x + 10)}} $[/tex]
⇒ [tex]$ 3^{4x} . 3^{-5} = \frac{1}{3^{6x.30}} $[/tex]
Also, [tex]$ a^x. a^{-y} = \frac{a^x}{a^{y}} $[/tex]
⇒ [tex]$ \frac{3^{4x}}{3^5} = \frac{1}{3^{6x}.3^{30}} $[/tex]
Cross-multiplying we get:
[tex]$ 3^{6x}.3^{30}.3^{4x} = 3^{5} $[/tex]
⇒ [tex]$ 3^{10x} = \frac{3^{5}}{3^{30}} = 3^{-25} $[/tex]
⇒[tex]$ 3^{10x} = 3^{-25} $[/tex]
Since the bases are same, the powers should be equal.
⇒ 10x = -25
⇒ x = [tex]$ \frac{-5}{2} $[/tex].
