Hei, kan noen vise/forklare meg hvordan disse likningene skam løse?
A) 8^x = 2^x+1
B) 5 × 6^x= 20 × 4^x
løse eksponentialikningene ved regning
Moderators: Aleks855, Gustav, Nebuchadnezzar, Janhaa, DennisChristensen, Emilga
[tex]8^x = 2^x+1[/tex] har ingen eksakt løsning, men kan løses vha. Newtons metode
[tex]f(x) = 2^{3x}-2^x-1 = 0[/tex]
[tex]f^\prime (x) = 3\cdot 2^{3x}\log{(2)}-2^{x}\log{(2)} = \log{(2)}\left(3\cdot 2^{3x}-2^x\right)[/tex]
[tex]x_{n+1} = x_n-\frac{f(x_n)}{f^\prime (x_n)}[/tex]
Som etter noen iterasjoner (jeg brukte 7) gir deg: [tex]x\approx 0.4057[/tex]
Er ligningen derimot [tex]8^x = 2^{x+1}[/tex] kan du gjøre sånn:
[tex]x\ln{(2^3)}=(x+1)\ln{(2)} \ \Rightarrow \ 3x\ln{(2)}-(x+1)\ln{(2)}=0 \ \Rightarrow \ \ln{(2)}\left(3x-x-1\right) = 0[/tex]
[tex]2x-1=0 \ \Rightarrow \ x = \frac{1}{2}[/tex]
B)
[tex]5\cdot 6^x = 20\cdot 4^x\ \Rightarrow \ \ln{(5)}+x\ln{(6)}=\ln{(20)}+x\ln{(4)} \ \Rightarrow \ x\left(\ln{(6)}-\ln{(4)}\right)+\ln{(5)}-\ln{(20)} = 0[/tex]
[tex]x\ln{\left(\frac{6}{4}\right)} + \ln{\left(\frac{5}{20}\right)} = 0[/tex]
[tex]x\ln{\left(\frac{3}{2}\right)} + \ln{\left(\frac{1}{4}\right)} = 0 \ \Rightarrow \ x = -\frac{\ln{\left(\frac{1}{4}\right)}}{\ln{\left(\frac{3}{2}\right)}}[/tex]
[tex]x = \frac{\ln{(4)}}{\ln{(3)}-\ln{(2)}} \approx 3.42[/tex]
[tex]f(x) = 2^{3x}-2^x-1 = 0[/tex]
[tex]f^\prime (x) = 3\cdot 2^{3x}\log{(2)}-2^{x}\log{(2)} = \log{(2)}\left(3\cdot 2^{3x}-2^x\right)[/tex]
[tex]x_{n+1} = x_n-\frac{f(x_n)}{f^\prime (x_n)}[/tex]
Som etter noen iterasjoner (jeg brukte 7) gir deg: [tex]x\approx 0.4057[/tex]
Er ligningen derimot [tex]8^x = 2^{x+1}[/tex] kan du gjøre sånn:
[tex]x\ln{(2^3)}=(x+1)\ln{(2)} \ \Rightarrow \ 3x\ln{(2)}-(x+1)\ln{(2)}=0 \ \Rightarrow \ \ln{(2)}\left(3x-x-1\right) = 0[/tex]
[tex]2x-1=0 \ \Rightarrow \ x = \frac{1}{2}[/tex]
B)
[tex]5\cdot 6^x = 20\cdot 4^x\ \Rightarrow \ \ln{(5)}+x\ln{(6)}=\ln{(20)}+x\ln{(4)} \ \Rightarrow \ x\left(\ln{(6)}-\ln{(4)}\right)+\ln{(5)}-\ln{(20)} = 0[/tex]
[tex]x\ln{\left(\frac{6}{4}\right)} + \ln{\left(\frac{5}{20}\right)} = 0[/tex]
[tex]x\ln{\left(\frac{3}{2}\right)} + \ln{\left(\frac{1}{4}\right)} = 0 \ \Rightarrow \ x = -\frac{\ln{\left(\frac{1}{4}\right)}}{\ln{\left(\frac{3}{2}\right)}}[/tex]
[tex]x = \frac{\ln{(4)}}{\ln{(3)}-\ln{(2)}} \approx 3.42[/tex]