Question 11)
Answer:
[tex]\log _2\left(63\right)-\log _2\left(9\right)=\log _2\left(7\right)=2.80[/tex]
Step-by-step explanation:
Given the expression
[tex]\log _2\left(63\right)-\log _2\left(9\right)[/tex]
[tex]\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)-\log _c\left(b\right)=\log _c\left(\frac{a}{b}\right)[/tex]
[tex]\log _2\left(63\right)-\log _2\left(9\right)=\log _2\left(\frac{63}{9}\right)[/tex]
[tex]=\log _2\left(\frac{63}{9}\right)[/tex]
[tex]\mathrm{Divide\:the\:numbers:}\:\frac{63}{9}=7[/tex]
[tex]=\log _2\left(7\right)[/tex]
[tex]=2.80[/tex]
Therefore,
[tex]\log _2\left(63\right)-\log _2\left(9\right)=\log _2\left(7\right)=2.80[/tex]
Question 12)
Answer:
[tex]\log \:_2\left(3\right)+\log \:_2\left(15\right)-\log \:_2\left(9\right)=\log \:_2\left(5\right)=2.32[/tex]
Step-by-step explanation:
Given the expression
[tex]\log _2\left(3\right)+\log _2\left(15\right)-\log _2\left(9\right)[/tex]
[tex]\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)+\log _c\left(b\right)=\log _c\left(ab\right)[/tex]
[tex]\log _2\left(3\right)+\log _2\left(15\right)=\log _2\left(3\cdot \:15\right)[/tex]
[tex]=\log _2\left(3\cdot \:15\right)-\log _2\left(9\right)[/tex]
[tex]\mathrm{Multiply\:the\:numbers:}\:3\cdot \:15=45[/tex]
[tex]=\log _2\left(45\right)-\log _2\left(9\right)[/tex]
[tex]\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)-\log _c\left(b\right)=\log _c\left(\frac{a}{b}\right)[/tex]
[tex]\log _2\left(45\right)-\log _2\left(9\right)=\log _2\left(\frac{45}{9}\right)[/tex]
[tex]=\log _2\left(\frac{45}{9}\right)[/tex]
[tex]\mathrm{Divide\:the\:numbers:}\:\frac{45}{9}=5[/tex]
[tex]=\log _2\left(5\right)[/tex]
[tex]=2.32[/tex]
Therefore,
[tex]\log \:_2\left(3\right)+\log \:_2\left(15\right)-\log \:_2\left(9\right)=\log \:_2\left(5\right)=2.32[/tex]
