Kan noen hjelpe meg med disse oppgavene?
Skriv uttrykkene enklere uten å bruke kalkulator:
1) e^2 * ln 4
2) e^2 ln 6 - ln 3
Takk!
To uttrykk!
Moderatorer: Aleks855, Gustav, Nebuchadnezzar, Janhaa, DennisChristensen, Emilga
Tja, ikke allverdens en kan gjøre med uttrykkene:
a)
e^2 * ln 4 = e^2 * 2ln2 = e^2 * e^(ln2) * ln2 = e^(2ln2) * ln2
e[sup]2[/sup] * ln4 = e[sup]2[/sup] * 2ln2 = e[sup]2[/sup]*e[sup]ln2[/sup]*ln2 = e[sup]2ln2[/sup]*ln2
b)
e^2 * ln6 - ln3 = e^2 * (ln3 + ln2) - ln3 = ln3 [e^2 -1] + e^2*ln2
e[sup]2[/sup]ln6 - ln3 = e[sup]2[/sup] (ln3+ln2) - ln3 = ln3 [e[sup]2[/sup]-1] + e[sup]2[/sup]ln2
MEN hvis du mente
a)
e^(2*ln4) = e^(ln4 *2) = (e^ln4)^2 = 4^2 = 16
e[sup]2*ln4[/sup] = e[sup]ln4 *2[/sup] = (e[sup]ln4[/sup])[sup]2[/sup] = 4[sup]2[/sup] = 16
b)
e^(2ln6-ln3) = e^(2ln6)*e^(-1*ln3) = (e^ln6)^2 * (e^ln3)^(-1) = 6^2 * 3^(-1) = 36 * (1/3) = 36/3 = 12
e[sup]2ln6-ln3[/sup] = e[sup]2ln6[/sup] e[sup]-ln3[/sup] = (e[sup]ln6[/sup])[sup]2[/sup] (e[sup]ln3[/sup])[sup]-1[/sup] = 6[sup]2[/sup] 3[sup]-1[/sup] = 36 * (1/3) = 36/3 = 12
a)
e^2 * ln 4 = e^2 * 2ln2 = e^2 * e^(ln2) * ln2 = e^(2ln2) * ln2
e[sup]2[/sup] * ln4 = e[sup]2[/sup] * 2ln2 = e[sup]2[/sup]*e[sup]ln2[/sup]*ln2 = e[sup]2ln2[/sup]*ln2
b)
e^2 * ln6 - ln3 = e^2 * (ln3 + ln2) - ln3 = ln3 [e^2 -1] + e^2*ln2
e[sup]2[/sup]ln6 - ln3 = e[sup]2[/sup] (ln3+ln2) - ln3 = ln3 [e[sup]2[/sup]-1] + e[sup]2[/sup]ln2
MEN hvis du mente
a)
e^(2*ln4) = e^(ln4 *2) = (e^ln4)^2 = 4^2 = 16
e[sup]2*ln4[/sup] = e[sup]ln4 *2[/sup] = (e[sup]ln4[/sup])[sup]2[/sup] = 4[sup]2[/sup] = 16
b)
e^(2ln6-ln3) = e^(2ln6)*e^(-1*ln3) = (e^ln6)^2 * (e^ln3)^(-1) = 6^2 * 3^(-1) = 36 * (1/3) = 36/3 = 12
e[sup]2ln6-ln3[/sup] = e[sup]2ln6[/sup] e[sup]-ln3[/sup] = (e[sup]ln6[/sup])[sup]2[/sup] (e[sup]ln3[/sup])[sup]-1[/sup] = 6[sup]2[/sup] 3[sup]-1[/sup] = 36 * (1/3) = 36/3 = 12