# Hvordan løse logaritmeulikheter

Når du løser logaritmeulikheter bruker du de vanlige reglene for ulikheter som du kjenner fra tidligere.

Regel

### Logaritmeulikheter

Det å opphøye høyre og venstre siden i et grunntall gjør at ulikheten forblir uendret. Altså, ingenting skjer med ulikhetstegnet! $\begin{array}{llllllll}\hfill \mathrm{lg}x&

Eksempel 1

Løs ulikheten $\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}\left(x-4\right)\le 8$ $\begin{array}{llll}\hfill \mathrm{ln}\phantom{\rule{-0.17em}{0ex}}\left(x-4\right)& \le 8\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {e}^{\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}\left(x-4\right)}& \le {e}^{8}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x-4& \le {e}^{8}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& \le {e}^{8}+4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& \le \text{}2984,96\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Eksempel 2

Løs ulikheten $\mathrm{lg}\phantom{\rule{-0.17em}{0ex}}\left(-2x+3\right)\ge 4$

$\begin{array}{llllll}\hfill \mathrm{lg}\phantom{\rule{-0.17em}{0ex}}\left(-2x+3\right)& \ge 4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {e}^{\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}\left(-2x+3\right)}& \ge {e}^{4}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -2x+3& \ge {e}^{4}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -2x& \ge {e}^{4}-3\phantom{\rule{2em}{0ex}}& \hfill & |:\phantom{\rule{-0.17em}{0ex}}\left(-2\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& \le \frac{{e}^{4}-3}{-2}=\frac{3-{e}^{4}}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ Du snur ulikheten siden du dividerer med et negativt tall. Du må ha likhetstegn i siste rad siden svaret kun skrives om slik at det ser penere ut!

Eksempel 3

Løs ulikheten $\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}\left(x-3\right)\le e$ $\begin{array}{llll}\hfill \mathrm{ln}\phantom{\rule{-0.17em}{0ex}}\left(x-3\right)& \le e\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {e}^{\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}\left(x-3\right)}& \le {e}^{e}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x-3& \le {e}^{e}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& \le {e}^{e}+3\approx \text{}18,15\text{}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Vil du vite mer?Registrer degDet er gratis!
Forrige oppslag
Hvordan løse eksponentialulikheter

# Hvordan løse logaritmeulikheter

Tall og måling
AlgebraGeometri
Statistikk og sannsynlighet
Funksjoner
Bevis