Question

Harga x yang memenuhi pertidaksamaan 2ˣ⁺¹+2⁵⁻ˣ ≤ 136 adalah ...
(A) -4 ≤ x ≤ 2 (B) 1/4 ≤ x ≤ 4 (C) -2 ≤ x ≤ 2 (D) -2 ≤ x ≤ 4 (E) -2 ≤ x ≤ 1/4

Answer 1

[tex]\bf{}^2\log\left(34-2\sqrt{285}\right)\ \le\ x\ \le\ {}^2\log\left(34+2\sqrt{285}\right)[/tex]atau[tex]\bf{-}2{,}08244...\ \lessapprox\ x\ \lessapprox 6{,}08244...[/tex] PembahasanEksponen dan Logaritma [tex]\begin{aligned}&2^{x+1}+2^{5-x}\ \le\ 136\\&{\Rightarrow\ }2\left(2^x\right)+\frac{32}{2^x}\ \le\ 136\\&{\quad}[\ \textsf{kalikan $2^x$}\ ]\\&{\Rightarrow\ }2\left(2^x\right)^2+32\ \le\ 136\left(2^x\right)\\&{\quad}[\ {\sf misalkan\ }u=2^x\ ]\\&{\Rightarrow\ }2u^2+32\ \le\ 136u\\&{\Rightarrow\ }2u^2-136u\ \le\ -32\\&{\quad}[\ \textsf{kedua ruas dibagi 2}\ ]\\&{\Rightarrow\ }u^2-68u\ \le\ -16\\&{\Rightarrow\ }u^2-68u+1156\ \le\ -16+1156\\&{\Rightarrow\ }(u-34)^2\ \le\ 1140\end{aligned}[/tex][tex]\begin{aligned}&{\quad}[\ \textsf{$u^n \le c$ dan $n$ genap }\Rightarrow -c^{\frac{1}{n}} \le u \le c^{\frac{1}{n}}\ ]\\&{\Rightarrow\ }{-}\sqrt{1140}\ \le\ u-34\ \le\ \sqrt{1140}\\&{\Rightarrow\ }34-\sqrt{1140}\ \le\ u\ \le\ 34+\sqrt{1140}\\&{\quad}[\ \sqrt{1140}=\sqrt{4\times285}=2\sqrt{285}\ ]\\&{\Rightarrow\ }34-2\sqrt{285}\ \le\ u\ \le\ 34+2\sqrt{285}\\&{\quad}[\ \textsf{substitusi $u$ kembali}\ ]\\&{\Rightarrow\ }34-2\sqrt{285}\ \le\ 2^x\ \le\ 34+2\sqrt{285}\end{aligned}[/tex][tex]\begin{aligned}&{\quad}[\ 2^x=a\ \Rightarrow\ x={}^2\log{a}\ ]\\&{\Rightarrow\ }\bf{}^2\log\left(34-2\sqrt{285}\right)\ \le\ x\ \le\ {}^2\log\left(34+2\sqrt{285}\right)\end{aligned}[/tex] Jika dihitung menggunakan kalkulator dengan pendekatan 5 angka di belakang koma:[tex]{\Rightarrow\ }\bf{-}2{,}08244...\ \lessapprox\ x\ \lessapprox 6{,}08244...[/tex]