Almira D
16 Januari 2023 01:03
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Almira D
16 Januari 2023 01:03
Pertanyaan
Tentukan nilai dari ((^(5)log 10)^(2)-(^(5)log 2)^(2))/(^(5)log√20)
6
1
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C. Salsa
Mahasiswa/Alumni Universitas Gajah Mada
03 Maret 2023 06:45
<p>Jawaban : 2<br><br>Ingat!<br>^a log b - ^a log c = ^a log (b/c) <br>^a log b + ^a log c = ^a log (bc) <br>^(a^(m)) log a^(n) = n/m<br><br>Sehingga </p><p>((^(5)log 10)^(2)-(^(5)log 2)^(2))/(^(5)log√20)<br>= ((^(5)log 10-^(5)log 2)(^(5)log 10+^(5)log 2))/(^(5)log√20)<br>= ((^(5)log 10/2)(^(5)log 10x2))/(^(5)log√20)<br>= ((^(5)log 5)(^(5)log 20)/(^(5)log 20^(1/2))<br>= ((1)(^(20^(1/2))log 20)<br>= ^(20^(1/2))log 20<br>= 1/(1/2)<br>= 2<br><br>Jadi, ((^(5)log 10)^(2)-(^(5)log 2)^(2))/(^(5)log√20) = 2</p>
Jawaban : 2
Ingat!
^a log b - ^a log c = ^a log (b/c)
^a log b + ^a log c = ^a log (bc)
^(a^(m)) log a^(n) = n/m
Sehingga
((^(5)log 10)^(2)-(^(5)log 2)^(2))/(^(5)log√20)
= ((^(5)log 10-^(5)log 2)(^(5)log 10+^(5)log 2))/(^(5)log√20)
= ((^(5)log 10/2)(^(5)log 10x2))/(^(5)log√20)
= ((^(5)log 5)(^(5)log 20)/(^(5)log 20^(1/2))
= ((1)(^(20^(1/2))log 20)
= ^(20^(1/2))log 20
= 1/(1/2)
= 2
Jadi, ((^(5)log 10)^(2)-(^(5)log 2)^(2))/(^(5)log√20) = 2
· 4.0 (2)
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