KUIS YUK!
Diketahui
[tex]\displaystyle\rm a = \:^5log2\\^{16}log(125)+\:^{625}log(8) = \frac{b(a^2+1)}{ac}[/tex]
Nilai c - b = ....![]()
Diketahui
[tex]\displaystyle\rm a = \:^5log2\\^{16}log(125)+\:^{625}log(8) = \frac{b(a^2+1)}{ac}[/tex]
Nilai c - b = ....
[tex] \rm^{16}log(125)+\:^{625}log(8) = \frac{b(a^2+1)}{ac}[/tex]
[tex] \rm^{ {2}^{4} }log( {5}^{3} )+\:^{ {5}^{4} }log( {2}^{3} ) = \frac{b(a^2+1)}{ac}[/tex]
[tex] \rm \frac{3}{4} \cdot \: ^{2}log(5)+\: \frac{3}{4} \cdot \: ^{5}log(2) = \frac{b(a^2+1)}{ac}[/tex]
[tex] \rm \frac{3}{4} \cdot \: \frac{1}{a} +\: \frac{3}{4} \cdot \:a = \frac{b(a^2+1)}{ac}[/tex]
[tex] \rm \frac{3}{4a} +\: \frac{3a}{4} = \frac{b(a^2+1)}{ac}[/tex]
[tex] \rm \frac{3}{4a} +\: \frac{3a^{2} }{4a} = \frac{b(a^2+1)}{ac}[/tex]
[tex] \rm \frac{3 + 3 {a}^{2} }{4a} = \frac{b(a^2+1)}{ac}[/tex]
[tex] \rm \frac{3 {a}^{2} + 3}{4a} = \frac{b(a^2+1)}{ac}[/tex]
[tex] \rm \frac{3( {a}^{2} + 1)}{4a} = \frac{b(a^2+1)}{ac}[/tex]
Diperoleh:
- b = 3
- c = 4
Jadi:
c - b
= 4 - 3
= 1
[answer.2.content]
