[tex]\displaystyle c) \:\:\frac{3^{2020}+3^{2021}+3^{2022}+3^{2023}}{120}=3^t\\\\\frac{3^{2020}+3^{2020+1}+3^{2020+2}+3^{2020+3}}{4\times3\times10}=3^t\\\\\frac{3^{2020}+3^{2020}(3^{1})+3^{2020}(3^{2})+3^{2020}(3^{3})}{2^3\times3\times5}=3^t\\\\\frac{3^{2020}(1+3^1+3^2+3^3)}{2^3\times3\times5}=3^t\\\\\frac{3^{2020-1}(1+3+9+27)}{2^3\times5}=3^t\\\\\frac{3^{2019}(40)}{2^3\times5}=3^t\\\\\frac{3^{2019}(8)}{2^3}=3^t\\\\\frac{3^{2019}(8)}{8}=3^t\\\\3^{2019}=3^t\\t=2019[/tex]
Jawaban:
[tex] \sf'비상' (Svt)[/tex]
Jawab:
(2.)
a. t = {-2, 2}
b. t = 902
c. t = 2019
Penjelasan dengan langkah-langkah:
(2.) Dicari nilai t untuk setiap nomor
[tex]\displaystyle a)\:\:\frac{(4^t)^t}{8}=32\\\\\frac{(2^2)^{t^2}}{2^3}=2^5\:\:\because\frac{n^a}{n^b}=n^{a-b}\therefore\\\\2^{2t^2-3}=2^5\\2t^2-3 = 5\\2t^2=5+3\\2t^2=8\\t^2=8/2\\t^2=4\\|t|=\sqrt{4}\\|t|=2\\t=\{-2,2\}[/tex]
[tex]\displaystyle b)\:\:2^{900}+4^{450}+8^{300}+16^{225}=2^t\\2^{900}+(2^2)^{450}+(2^3)^{300}+(2^4)^{225}=2^t\\2^{900}+2^{2\times450}+2^{3\times300}+2^{4\times225}=2^t\\2^{900}+2^{900}+2^{900}+2^{900}=2^t\\4(2^{900})=2^t\\2^2\times(2^{900})=2^t\:\:\because n^a\times n^b = n^{a+b}\therefore\\2^{2+900}=2^t\\2^{902}=2^t\\t=902[/tex]
[tex]\displaystyle c) \:\:\frac{3^{2020}+3^{2021}+3^{2022}+3^{2023}}{120}=3^t\\\\\frac{3^{2020}+3^{2020+1}+3^{2020+2}+3^{2020+3}}{4\times3\times10}=3^t\\\\\frac{3^{2020}+3^{2020}(3^{1})+3^{2020}(3^{2})+3^{2020}(3^{3})}{2^3\times3\times5}=3^t\\\\\frac{3^{2020}(1+3^1+3^2+3^3)}{2^3\times3\times5}=3^t\\\\\frac{3^{2020-1}(1+3+9+27)}{2^3\times5}=3^t\\\\\frac{3^{2019}(40)}{2^3\times5}=3^t\\\\\frac{3^{2019}(8)}{2^3}=3^t\\\\\frac{3^{2019}(8)}{8}=3^t\\\\3^{2019}=3^t\\t=2019[/tex]
(xcvi)