Question

Daca log in baza 2 din 3 = a atunci cu cat egal log in baza 18 din 24?

Answer 1

Răspuns:

$log_{18}24 = \frac{3 + a}{1 + 2 a}$

Explicație pas cu pas:

Datele problemei:

$log_23 = a$

Ce se cere:

$S\u{a} \ se \ scrie\ num\u{a}rul\ log_{18}24 \ \^{i}n\ func\c{t}ie \de\ a.$

Rezolvare:

$\textbf{Formule\ utilizate\ \^{i}n\ cadrul\ rezolv\u{a}rii:}\\\emph{Mai jos se reg\u{a}sesc\ c\^{a}teva\ dintre\ propriet\u{a}\c{t}ile logaritmilor.}$

• $log_a a = 1$

• $log_aa^m = m$

• $log_a b^m = mlog_ab$

• $log_ab = \frac{log_cb}{log_ca} \ \ ( denumit\u{a}\ formula\ pentru\ schimbare\ a\ bazei)$

• $log_a xy = log_ax + log_a y$

$Voi\ transforma\ log_{18}24 \ \^{i}n\ baz\u{a}\ 2 .$

$log_{18} 24 = \frac{log_224}{log_218}$

$\\Scriu\ numerele\ 24\ \c{s}i\ 18\ ca\ produs\ de\ 2\ la\ putere\ \^{i}nmul\c{t}it\ cu\ alt\ num\u{a}r.$$24 = 2^3 * 3\\18 = 2 * 3^2$

$log_{18}24 = \frac{log_2 (2^3 * 3)}{log_2(2*3^2)}$

$Transform\ \^{i}n\ sum\u{a}\ de\ logaritmi:\\log_{18}24 = \frac{log_2 (2^3 * 3)}{log_2(2*3^2)} = \frac{log_2 2^3 + log_23}{log_22+log_23^2}$

$Folosesc\ a\ treia\ proprietate\ men\c{t}ionat\u{a}\ mai\ sus:$

$log_{18}24 =\frac{3log_22+log_23}{log_22+2log_23}$

$Cum\ log_2 2 = 1\ \ \c{s}i\ identific\^{a}nd\ expresia\ lui\ a , \ rela\c{t}ia\ de\ mai\ sus\ se\ rescrie\ astfel:\\\\\log_{18}24 = \frac{3 * 1 + a}{1 + 2 * a}\\\\log_{18}24 = \frac{3 + a}{1 + 2 a}$