Question

Preciso de ajuda com logaritmos.
Questão: Sendo log 2 = a e log 3 = b, calcule:
a) log 15
b) log 36
c) log 0,75

Answer 1

Olá Sara,

dadas as propriedades de log..

$\boxed{log(ab)~\to~log(a\cdot b)~\to~loga+logb}\\\\ \boxed{log\left( \dfrac{a}{b}\right)~\to~loga-logb}\\\\ \boxed{log(b)^a~\to~a\cdot log(b)}\\\\ \boxed{log_aa=1}$

$log15=log(3\cdot5)\\ log15=log3+log5\\ log15=log3+log\left( \dfrac{10}{2}\right)\\\\ log15=log3+(log10-log2)\\\\ \Large\boxed{\boxed{log15=b+(1-a)}}$

_____________

$log36=log(2^2\cdot3^2)\\ log36=log(2)^2+log(3)^2\\ log36=2\cdot log2+2\cdot log3\\ log36=2\cdot a+2\cdot b\\\\ \Large\boxed{\boxed{log36=2\cdot(a+b)}}$

_____________

$log0,75=log\left( \dfrac{75}{100}\right)\\\\ log0,75=log75-log100\\ log0,75=log(3\cdot5^2)-log(10)^2\\ log0,75=log3+log(5)^2-2\cdot log10\\ log0,75=log3+2\cdot log5-2\cdot log10\\ log0,75=log3+2\cdot log\left( \dfrac{10}{2}\right)-2\cdot log10\\ log0,75=[log3+2\cdot (log10-log2)-2\cdot log10]\\ log0,75=[b+2\cdot(1-a)-2\cdot1]\\ log0,75=b+2-2a-2\\\\ \huge\boxed{\boxed{log0,75=b-2a~~ou~~-2a+b}}$

Tenha ótimos estudos ;D