Correction to the last assignment (1)(1/(81^(x−2) )) = 27^(1−x) (1/3^(4(x−2)) )=3^(3(1−x)) (1/3^(4x−8) )=3^(3−3x) 3^(−4x+8) =3^(3−3x) −4x+8=3−3x C.L.T −4x+3x=3−8 −x=−5 x=5 (2)9^x =(1/(729)) 9^x =(1/3^6 ) 3^(2x) =3^(−6) 2x=−6 x=((−6)/2) x=−3 (3)16^x =0.125 2^(4x) =((125)/(1000)) 2^(4x) =(1/8) 2^(4x) =2^(−3) 4x=−3 x=((−3)/4) (4)a^(1/2) =4 a=


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Question Number 120241 by olalekanoriyomi last updated on 30/Oct/20

$C o r r e c t i o n t o t h e l a s t a s s i g n m e n t$ $(1) \frac{1}{81^{x - 2}} = 27^{1 - x}$ $\frac{1}{3^{4 (x - 2)}} = 3^{3 (1 - x)}$ $\frac{1}{3^{4 x - 8}} = 3^{3 - 3 x}$ $3^{- 4 x + 8} = 3^{3 - 3 x}$ $- 4 x + 8 = 3 - 3 x$ $C . L . T$ $- 4 x + 3 x = 3 - 8$ $- x = - 5$ $x = 5$ $(2) 9^{x} = \frac{1}{729}$ $9^{x} = \frac{1}{3^{6}}$ $3^{2 x} = 3^{- 6}$ $2 x = - 6$ $x = \frac{- 6}{2}$ $x = - 3$ $(3) 16^{x} = 0.125$ $2^{4 x} = \frac{125}{1000}$ $2^{4 x} = \frac{1}{8}$ $2^{4 x} = 2^{- 3}$ $4 x = - 3$ $x = \frac{- 3}{4}$ $(4) a^{1 / 2} = 4$ $a =$
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