\frac{ \sqrt[3]{4 ^{5-x} } }{8} = \frac{1}{ 2^{2x+1} } nilai x adalah....
Matematika
LuluuuuuFF
Pertanyaan
\frac{ \sqrt[3]{4 ^{5-x} } }{8} = \frac{1}{ 2^{2x+1} } nilai x adalah....
1 Jawaban
-
1. Jawaban Takamori37
[tex]\displaystyle \frac{ \sqrt[3]{4 ^{5-x} } }{8} = \frac{1}{ 2^{2x+1} } \\ \frac{4^{\frac{5-x}{3}}}{8}=2^{-(2x+1)} \\ 8^{-1}.2^{\frac{10-2x}{3}}=2^{-2x-1} \\ 2^{-3}.2^{\frac{10-2x}{3}}=2^{-2x-1} \\ 2^{\frac{10-2x}{3}-3}}=2^{-2x-1} \\ 2^{\frac{1-2x}{3}}}=2^{-2x-1} \\ \frac{1-2x}{3}=-2x-1 \\ 1-2x=-6x-3 \\ 4x=-4 \\ x=-1[/tex]