Pertanyaan

Tentukan himpunan penyelesaian persamaan berikut. c. ( 2 1 ) 2 x + 1 = 16 2 4 x − 1 ( 0 , 125 )

Tentukan himpunan penyelesaian persamaan berikut.

c. $(\frac{1}{2})^{2 x + 1} = \frac{2 ^{4 x - 1} ( 0 , 125 )}{16}$

N. Puspita

Master Teacher

Jawaban terverifikasi

Jawaban

himpunan penyelesaian persamaan tersebut adalah .

himpunan penyelesaian persamaan tersebut adalah open curly brackets 3 over 4 close curly brackets

Pembahasan

Ingat kembali bentuk persamaan eksponen berikut: Diketahuipersamaan . Maka: Jadi, himpunan penyelesaian persamaan tersebut adalah .

Ingat kembali bentuk persamaan eksponen berikut:

a to the power of f open parentheses x close parentheses end exponent equals a to the power of g open parentheses x close parentheses end exponent comma space a greater than 0 space dan space a not equal to 1 rightwards arrow f open parentheses x close parentheses equals g open parentheses x close parentheses

Diketahui persamaan $open parentheses 1 half close parentheses to the power of 2 x plus 1 end exponent equals square root of fraction numerator 2 to the power of 4 x minus 1 end exponent open parentheses 0 comma 125 close parentheses over denominator 16 end fraction end root$ . Maka:

$table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 1 half close parentheses to the power of 2 x plus 1 end exponent end cell equals cell square root of fraction numerator 2 to the power of 4 x minus 1 end exponent open parentheses 0 comma 125 close parentheses over denominator 16 end fraction end root end cell row cell open parentheses 2 to the power of negative 1 end exponent close parentheses to the power of 2 x plus 1 end exponent end cell equals cell square root of fraction numerator 2 to the power of 4 x minus 1 end exponent open parentheses begin display style fraction numerator 125 over denominator 1.000 end fraction end style close parentheses over denominator 2 to the power of 4 end fraction end root end cell row cell 2 to the power of negative 2 x minus 1 end exponent end cell equals cell square root of fraction numerator 2 to the power of 4 x minus 1 end exponent open parentheses begin display style 1 over 8 end style close parentheses over denominator 2 to the power of 4 end fraction end root end cell row cell 2 to the power of negative 2 x minus 1 end exponent end cell equals cell square root of fraction numerator 2 to the power of 4 x minus 1 end exponent open parentheses begin display style 1 over 2 cubed end style close parentheses over denominator 2 to the power of 4 end fraction end root end cell row cell 2 to the power of negative 2 x minus 1 end exponent end cell equals cell square root of fraction numerator 2 to the power of 4 x minus 1 end exponent open parentheses 2 to the power of negative 3 end exponent close parentheses over denominator 2 to the power of 4 end fraction end root end cell row cell 2 to the power of negative 2 x minus 1 end exponent end cell equals cell square root of 2 to the power of 4 x minus 1 plus open parentheses negative 3 close parentheses end exponent over 2 to the power of 4 end root end cell row cell 2 to the power of negative 2 x minus 1 end exponent end cell equals cell square root of 2 to the power of 4 x minus 4 end exponent over 2 to the power of 4 end root end cell row cell 2 to the power of negative 2 x minus 1 end exponent end cell equals cell square root of 2 to the power of 4 x minus 4 minus 4 end exponent end root end cell row cell 2 to the power of negative 2 x minus 1 end exponent end cell equals cell square root of 2 to the power of 4 x minus 8 end exponent end root end cell row cell 2 to the power of negative 2 x minus 1 end exponent end cell equals cell 2 to the power of fraction numerator 4 x minus 8 over denominator 2 end fraction end exponent end cell row cell 2 to the power of negative 2 x minus 1 end exponent end cell equals cell 2 to the power of 2 x minus 4 end exponent end cell row cell negative 2 x minus 1 end cell equals cell 2 x minus 4 end cell row cell negative 2 x minus 2 x end cell equals cell negative 4 plus 1 end cell row cell negative 4 x end cell equals cell negative 3 end cell row x equals cell fraction numerator negative 3 over denominator negative 4 end fraction end cell row x equals cell 3 over 4 end cell end table$

Jadi, himpunan penyelesaian persamaan tersebut adalah open curly brackets 3 over 4 close curly brackets .