Question

Selesaikan soal ini dengan strategi yang menurut anda termudah
[tex]\displaystyle \sqrt{^2\log 3~^2\log 12~^2\log 48~^2\log 192+16}-^2\log 12~^2\log 48+10[/tex]

Answer 1

Verified answer

Jawab:
= 6

Penjelasan:
[tex]\displaystyle\sqrt{^2\log 3~^2\log 12~^2\log 48~^2\log 192+16}-^2\log 12~^2\log 48+10\\\\=\sqrt{(^2\log 3~^2\log (2^2\cdot3)~^2\log (2^4\cdot3)~^2\log (2^6\cdot3))+16}-^2\log (2^2\cdot3)~^2\log (2^4\cdot3)+10\\\\=\sqrt{(^2\log 3~(^2\log (2^2)+^2\log3)~(^2\log (2^4)+^2\log3)~(^2\log (2^6)+^2\log3))+16}-(^2\log (2^2)+^2\log3)~(^2\log (2^4)+^2\log3)+10\\\\=\sqrt{(^2\log 3~(2+^2\log3)~(4+^2\log3)~(6+^2\log3))+16}-(2+^2\log3)~(4+^2\log3)+10[/tex]Pakai permisalan. ²log 3 = u
[tex]\displaystyle=\sqrt{(u~(2+u)~(4+u)~(6+u))+16}-(2+u)~(4+u)+10\\\\=\sqrt{(2u+u^2)~(24+10u+u^2)+16}-(8+6u+u^2)+10\\\\=\sqrt{48u+20u^2+2u^3+24u^2+10u^3+u^4+16}-8-6u-u^2+10\\\\=\sqrt{u^4+12u^3+44u^2+48u+16}-8-6u-u^2+10[/tex]

Maka
[tex]\displaystyle\sqrt{u^4+12u^3+44u^2+48u+16}-8-6u-u^2+10\\\\=\sqrt{(u^2)^2+12u^3+44u^2+2(24u)+4^2}-8-6u-u^2+10\\\\=\sqrt{(u^2)^2+\left[12u^3+44u^2\right]+2(4)(6u)+4^2}-8-6u-u^2+10\\\\=\sqrt{(u^2)^2+\left[2(6u^3)+44u^2\right]+2(4)(6u)+4^2}-8-6u-u^2+10\\\\=\sqrt{(u^2)^2+2(u^2)(6u)+\left[44u^2\right]+2(4)(6u)+4^2}-8-6u-u^2+10\\\\=\sqrt{(u^2)^2+2(u^2)(6u)+\left[36u^2+8u^2\right]+2(4)(6u)+4^2}-8-6u-u^2+10\\\\=\sqrt{(u^2)^2+2(u^2)(6u)+\left[(6u)^2+2(4)u^2\right]+2(4)(6u)+4^2}-8-6u-u^2+10[/tex]

[tex]\displaystyle=\sqrt{(u^2)^2+2(u^2)(6u)+(6u)^2+2(4)u^2+2(4)(6u)+4^2}-8-6u-u^2+10\\\\=\sqrt{(u^2)^2+(6u)^2+4^2+2(u^2)(6u)+2(u^2)(4)+2(6u)(4)}-8-6u-u^2+10\\\\=\sqrt{(u^2)^2+(6u)^2+4^2+2((u^2)(6u)+(u^2)(4)+(6u)(4))}-8-6u-u^2+10[/tex]
Polanya mulai tampak
[tex]\displaystyle\because (x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)\therefore\\\\\sqrt{(u^2)^2+(6u)^2+4^2+2((u^2)(6u)+(6u)(4)+(u^2)(4))}-8-6u-u^2+10\\\\=\sqrt{(u^2+6u+4)^2}-8-6u-u^2+10[/tex]
=      u² + 6u  + 4   -   8  - 6u - u²  + 10
=  (u² - u²) + (6u - 6u) + 4 - 8 + 10
=  4 - 8 + 10
=  - (8-4) + 10
=  - 4 + 10
= 10 - 4
= 6
(xcvi)