Diberikan bahwa , cari nilai . Given that , find the value of .

Pertanyaan

Diberikan bahwa $3 squared equals 9 to the power of k$ , cari nilai $k$ .
Given that $3 squared equals 9 to the power of k$ , find the value of $k$ .

1 $space$
2 $space$
3 $space$
4 $space$

Pembahasan Video:

Pembahasan Soal:

$table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 3 squared equals 9 to the power of k 3 squared equals open parentheses 3 squared close parentheses to the power of k 3 squared equals 3 to the power of 2 k end exponent 2 equals 2 k k equals 2 over 2 k equals 1 end cell end table$

Jadi, jawaban yang tepat adalah A.

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Dijawab oleh:

N. Puspita

Terakhir diupdate 13 September 2021

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Jika , maka nilai adalah ....

Pembahasan Soal:

Diketahui $begin mathsize 14px style square root of 3 square root of 9 square root of 27 end root end root equals 3 to the power of x over y end exponent end style$ . Akan dicari terlebih dahulu nilai $begin mathsize 14px style x end style$ dan $begin mathsize 14px style y end style$ .

$table attributes columnalign right center left columnspacing 0px end attributes row cell square root of size 14px 3 square root of size 14px 9 square root of size 14px 27 end root end root end cell size 14px equals cell square root of size 14px 3 square root of size 14px 3 to the power of size 14px 2 square root of size 14px 3 to the power of size 14px 3 end root end root end root end cell row blank size 14px equals cell square root of size 14px 3 square root of size 14px 3 to the power of size 14px 2 begin mathsize 14px style left parenthesis 3 to the power of 3 over 2 end exponent right parenthesis end style end root end root end cell row blank size 14px equals cell square root of size 14px 3 square root of size 14px 3 to the power of size 14px 2 size 14px plus size 14px 3 over size 14px 2 end exponent end root end root end cell row blank size 14px equals cell square root of size 14px 3 square root of size 14px 3 to the power of size 14px 7 over size 14px 2 end exponent end root end root end cell row blank size 14px equals cell square root of size 14px 3 begin mathsize 14px style left parenthesis 3 to the power of 7 over 2 end exponent right parenthesis end style to the power of size 14px 1 over size 14px 2 end exponent end root end cell row blank size 14px equals cell square root of size 14px 3 to the power of size 14px 1 size 14px plus size 14px 7 over size 14px 4 end exponent end root end cell row blank size 14px equals cell begin mathsize 14px style left parenthesis 3 to the power of 11 over 4 end exponent right parenthesis end style to the power of size 14px 1 over size 14px 2 end exponent end cell row blank size 14px equals cell size 14px 3 to the power of size 14px 11 over size 14px 8 end exponent end cell end table$

Diperoleh $square root of size 14px 3 square root of size 14px 9 square root of size 14px 27 end root end root size 14px equals size 14px 3 to the power of size 14px x over size 14px y end exponent size 14px equals size 14px 3 to the power of size 14px 11 over size 14px 8 end exponent$ maka nilai $begin mathsize 14px style x equals 11 end style$ dan $begin mathsize 14px style y equals 8 end style$ .

Kemudian didapat nilai $begin mathsize 14px style x plus y equals 11 plus 8 equals 19 end style$

Jadi, jawaban yang benar adalah D.

Roboguru

Nilai yang memenuhi penyelesaian dari adalah ....

Pembahasan Soal:

$Error converting from MathML to accessible text.$

abaikan angka 2 sehingga fokus pada pangkat.

$undefined$

Jadi, jawaban yang tepat adalah A.

Roboguru

Tentukan himpunan penyelesaian dari : b. Catatan : Tunjukkan sifat/rumus yang digunakan dari penyelesaian tersebut.

Pembahasan Soal:

Sifat bilangan berpangkat yang digunakan:
$1. space open parentheses a to the power of m close parentheses to the power of n equals a to the power of m n end exponent 2. space a to the power of negative m end exponent equals 1 over a to the power of m$

Maka,

$table attributes columnalign right center left columnspacing 0px end attributes row cell 9 to the power of x minus 6 end exponent end cell equals cell 1 over 27 square root of 3 end cell row cell 3 to the power of 2 left parenthesis x minus 6 right parenthesis end exponent end cell equals cell 3 to the power of negative 3 plus bevelled 1 half end exponent end cell row cell 2 x minus 12 end cell equals cell negative 5 over 2 end cell row cell 2 x end cell equals cell negative 5 over 2 plus 12 end cell row cell 2 x end cell equals cell 19 over 2 end cell row x equals cell 19 over 4 end cell end table$

Jadi, nilai x yang memenuhi adalah $begin mathsize 14px style x equals 19 over 4. end style$

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Jika , maka nilai

Pembahasan Soal:

$begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell cube root of y square root of y end root end cell equals 5 row cell cube root of y square root of y end root end cell equals cell cube root of 125 end cell row cell cube root of y square root of y end root end cell equals cell cube root of 25 cross times 5 end root end cell row cell cube root of y square root of y end root end cell equals cell cube root of 25 square root of 25 end root end cell row y equals 25 end table end style$

Jadi, jawaban yang tepat adalah C.

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Jika , ialah...

Pembahasan Soal:

Ingat sifat eksponen :

$1 over a equals a to the power of negative 1 end exponent open parentheses a to the power of m close parentheses to the power of n equals a to the power of m cross times n end exponent$

berdasarkan sifat tersebut, maka

$table attributes columnalign right center left columnspacing 0px end attributes row cell 4 to the power of negative 3 over 2 end exponent end cell equals cell open parentheses 1 half close parentheses to the power of n end cell row cell open parentheses 2 squared close parentheses to the power of blank to the power of negative 3 over 2 end exponent end exponent end cell equals cell open parentheses 2 to the power of negative 1 end exponent close parentheses to the power of n end cell row cell 2 to the power of negative 3 end exponent end cell equals cell 2 to the power of negative n end exponent end cell row cell negative 3 end cell equals cell negative n end cell row n equals 3 end table$

Jadi, jawaban yang tepat adalah D.

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