Dear student, If the question is to solve for x then refer to the solution below:- We have,432x+12=132⇒413×2x+12=125⇒2213×2x+12=125⇒223×2x+12=2-5 [using amn=amn and a-m=1am]Equating the powers since the base is same⇒23×2x+12=-5⇒2x+12=-5×32⇒2x=-152-12⇒2x=-15-12⇒2x=-162⇒2x=-8⇒x=-82⇒x=-4thus the value of x is -4 Regards

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Dear student,

If the question is to solve for x then refer to the solution below:-

W e h a v e, {(\sqrt[3]{4})}^{2 x + \frac{1}{2}} = \frac{1}{32} \Rightarrow {(4)}^{\frac{1}{3} \times} (^{2 x + \frac{1}{2}}) = {(\frac{1}{2})}^{5} \Rightarrow {(2^{2})}^{\frac{1}{3} \times} (^{2 x + \frac{1}{2}}) = {(\frac{1}{2})}^{5} \Rightarrow {(2)}^{^{\frac{2}{3} \times} (^{2 x + \frac{1}{2}})} = {(2)}^{- 5} . . . [u \sin g {(a^{m})}^{n} = a^{m n} a n d {(a)}^{- m} = {(\frac{1}{a})}^{m}] E q u a t i n g t h e p o w e r s \sin c e t h e b a s e i s s a m e \Rightarrow \frac{2}{3} \times (2 x + \frac{1}{2}) = - 5 \Rightarrow 2 x + \frac{1}{2} = \frac{- 5 \times 3}{2} \Rightarrow 2 x = \frac{- 15}{2} - \frac{1}{2} \Rightarrow 2 x = \frac{- 15 - 1}{2} \Rightarrow 2 x = \frac{- 16}{2} \Rightarrow 2 x = - 8 \Rightarrow x = \frac{- 8}{2} \Rightarrow x = - 4 t h u s t h e v a l u e o f x i s - 4

Regards

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