Question

1)sabendo que log 2 = a e log3 = b, calcule, em função de a e b:
a)log 6
b)log 1,5
c)log 5
d) log 30
e)log1/4
f)log 72
g)log0,3
h)log∛1,8
i)log 0,024
j)log 0,75

Answer 1

a) $log6} \\\\ log2 \cdot 3 \\\\ log2+log3 \\\\ \boxed{\boxed{a+b}}$


b) $log1,5 \\\\ log\frac{3}{2} \\\\ log3-log2 \\\\ \boxed{\boxed{b-a}}$


c) $log5 \\\\ log\frac{10}{2} \\\\ log10-log2 \\\\ \boxed{\boxed{1-a}}$


d) $log30 \\\\ log2\cdot 3 \cdot 5 \\\\ log2+log3+log5 \\\\ a+b+(1-a) \\\\ a+b+1-a \\\\ \boxed{\boxed{b+1}}$


e) $log\frac{1}{4} \\\\ log1+log2^{2} \\\\ 0+2 \cdot log2 \\\\ \boxed{\boxed{2a}}$


f)$log72 \\\\ log3^{2} \cdot 2^{2} \cdot 2 \\\\ log3^{2}+log2^{2}+log2 \\\\ 2 \cdot log3 + 2 \cdot log2+log2 \\\\ 2b+2a+a \\\\ \boxed{\boxed{2b+3a}}$ 


g) $log0,3 \\\\ log\frac{3}{10} \\\\ log3-log10 \\\\ \boxed{\boxed{b-1}}$


h) $log \sqrt[3]{1,8} \\\\ log1,8^{\frac{1}{3}} \\\\ \frac{1}{3} \cdot (log\frac{18}{10}) \\\\ \frac{1}{3} \cdot (log18-log10) \\\\ \frac{1}{3} \cdot (log 3^{2} \cdot 2 - 1) \\\\ \frac{1}{3} \cdot (2log3+log2-1) \\\\ \frac{1}{3} \cdot (2b+a-1) \\\\ \boxed{\boxed{\frac{2b}{3}+\frac{a}{3}-\frac{1}{3}}}$


i) $log0,024 \\\\ log\frac{24^{\div 8}}{1000^{\div 8}} \\\\ log\frac{3}{125} \\\\ log3-log125 \\\\ b-log5^{3} \\\\ b-3 \cdot log5 \\\\ b-3 \cdot (1-a) \\\\ \boxed{\boxed{b-3+3a}}$


j) $log0,75 \\\\ log\frac{75}{100} \\\\ log\frac{3}{4} \\\\ log3-log4 \\\\ b-log2^{2} \\\\ b-2log2 \\\\ \boxed{\boxed{b-2a}}$