jika x = 3log2, maka √(27^x + 8 . 3^x+2) = ....?

x = 3 log 2
3^x = 2

√(27^x + 8 . 3^x + 2)
= √[(3^3)^x + 8 . 3^x + 2]
= √[(3^x)^3 + 8 . 3^x + 2]

Selanjutnya, ganti (substitusi) 3^x dengan 2
Maka diperoleh:

= √(2^3 + 8 . 2 + 2)
= √(8 + 8 . 2 + 2)
= √(8 + 16 + 2)
=√26


Semoga membantu :)

[tex]
x= 3 log(2). = > {3}^{x} = 2 \\ = > \sqrt{ {27}^{x} + 8 \: . \: {3}^{x + 2} } \\ = \sqrt{ {( {3}^{x}) }^{3} + 8 \: . \: {3}^{x} \times 9} \\ = \sqrt{ { ({3}^{x} )}^{3} + 72 \times {3}^{x} } \\ = \sqrt{ {2}^{3} + 72 \times 2} \\ = \sqrt{8 + 144} \\ = \sqrt{152} \\ = 2 \sqrt{38} [/tex]
Semoga Membantu √