Răspuns:
[tex]log_{18}24 = \frac{3 + a}{1 + 2 a}\\[/tex]
Explicație pas cu pas:
Datele problemei:
[tex]log_23 = a[/tex]
Ce se cere:
[tex]S\u{a} \ se \ scrie\ num\u{a}rul\ log_{18}24 \ \^{i}n\ func\c{t}ie \de\ a.[/tex]
Rezolvare:
[tex]\textbf{Formule\ utilizate\ \^{i}n\ cadrul\ rezolv\u{a}rii:}\\\emph{Mai jos se reg\u{a}sesc\ c\^{a}teva\ dintre\ propriet\u{a}\c{t}ile logaritmilor.}[/tex]
- [tex]log_a b^m = mlog_ab[/tex]
- [tex]log_ab = \frac{log_cb}{log_ca} \ \ ( denumit\u{a}\ formula\ pentru\ schimbare\ a\ bazei)[/tex]
- [tex]log_a xy = log_ax + log_a y[/tex]
[tex]Voi\ transforma\ log_{18}24 \ \^{i}n\ baz\u{a}\ 2 .[/tex]
[tex]log_{18} 24 = \frac{log_224}{log_218}[/tex]
[tex]\\Scriu\ numerele\ 24\ \c{s}i\ 18\ ca\ produs\ de\ 2\ la\ putere\ \^{i}nmul\c{t}it\ cu\ alt\ num\u{a}r.[/tex][tex]24 = 2^3 * 3\\18 = 2 * 3^2\\[/tex]
[tex]log_{18}24 = \frac{log_2 (2^3 * 3)}{log_2(2*3^2)}[/tex]
[tex]Transform\ \^{i}n\ sum\u{a}\ de\ logaritmi:\\log_{18}24 = \frac{log_2 (2^3 * 3)}{log_2(2*3^2)} = \frac{log_2 2^3 + log_23}{log_22+log_23^2}[/tex]
[tex]Folosesc\ a\ treia\ proprietate\ men\c{t}ionat\u{a}\ mai\ sus:[/tex]
[tex]log_{18}24 =\frac{3log_22+log_23}{log_22+2log_23}[/tex]
[tex]Cum\ log_2 2 = 1\ \ \c{s}i\ identific\^{a}nd\ expresia\ lui\ a , \ rela\c{t}ia\ de\ mai\ sus\ se\ rescrie\ astfel:\\\\\log_{18}24 = \frac{3 * 1 + a}{1 + 2 * a}\\\\log_{18}24 = \frac{3 + a}{1 + 2 a}[/tex]
