Question

Jika [tex]\displaystyle 2^x=2-\sqrt{3}[/tex] maka nilai dari [tex]\displaystyle ^{2+\sqrt{3}}\log 4^x[/tex] adalah ...

Answer 1

Jawaban:

maaf ya kak kalo salah , semoga membantu ya kak

Answer 2

Verified answer

Jika [tex]2^x=2-\sqrt{3}[/tex] maka nilai dari [tex]{}^{2+\sqrt{3}}\log4^x[/tex] adalah –2.
 

Penjelasan dengan langkah-langkah:

Logaritma

[tex]\begin{aligned}&{}^{2+\sqrt{3}}\log4^x\\&{=\ }\frac{\log4^x}{\log\left(2+\sqrt{3}\right)}\\&{=\ }\frac{\log\left(2^x\right)^2}{\log\left(2+\sqrt{3}\right)}\\&{=\ }2\cdot\frac{\log2^x}{\log\left(2+\sqrt{3}\right)}\\&{=\ }2\cdot\frac{\log\left(2-\sqrt{3}\right)}{\log\left(2+\sqrt{3}\right)}\\&{\quad...\ }\left[\begin{aligned}&\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)=4-3=1\\&\Rightarrow 2-\sqrt{3}=\frac{1}{2+\sqrt{3}}=\left(2+\sqrt{3}\right)^{-1}\end{aligned}\right]\end{aligned}[/tex]
[tex]\begin{aligned}&{=\ }2\cdot\frac{\log\left(2+\sqrt{3}\right)^{-1}}{\log\left(2+\sqrt{3}\right)}\\&{=\ }2\cdot(-1)\cdot\frac{\cancel{\log\left(2+\sqrt{3}\right)}}{\cancel{\log\left(2+\sqrt{3}\right)}}\\&{=\ }\boxed{\bf{-2}}\quad\blacksquare\end{aligned}[/tex]