log8+log25+2log3-log18
Answers
Answer:
2
Step-by-step explanation:
To solve this problem, first you have to used log formula.
Given:
log 8+ log 25+2 log 3- log 18
Solutions:
Used log rules.
\Large\boxed{\textnormal{LOG RULES FORMULA}}
LOG RULES FORMULA
\displaystyle \log _c\left(a\right)+\log _c\left(b\right)=\log _c\left(ab\right)log
c
(a)+log
c
(b)=log
c
(ab)
\displaystyle\log _{10}\left(8\right)+\log _{10}\left(25\right)=\log _{10}\left(8\cdot \:25\right)log
10
(8)+log
10
(25)=log
10
(8⋅25)
\displaystyle \log _{10}\left(8\cdot \:25\right)+2\log _{10}\left(3\right)-\log _{10}\left(18\right)log
10
(8⋅25)+2log
10
(3)−log
10
(18)
Multiply the numbers from left to right.
\displaystyle 8*25=2008∗25=200
\displaystyle \log _{10}\left(200\right)+2\log _{10}\left(3\right)-\log _{10}\left(18\right)log
10
(200)+2log
10
(3)−log
10
(18)
\Large\boxed{{a\log _c\left(b\right)=\log _c\left(b^a\right)}}
alog
c
(b)=log
c
(b
a
)
\displaystyle 2\log _{10}\left(3\right)=\log _{10}\left(3^2\right)2log
10
(3)=log
10
(3
2
)
$$\displaystyle \log _{10}(200)+\log _{10}\(3^2)-\log _{10}(18)$$
$$\displaystyle \log _{10}\left(200\right)+\log _{10}\left(3^2\right)=\log _{10}\left(3^2\cdot \:200\right)$$
$$\displaystyle \log _{10}\left(3^2\cdot \:200\right)-\log _{10}\left(18\right)$$
Solve.
Multiply.
$$\displaystyle 3^2*200=1800$$
$$\displaystyle \log _{10}\left(1800\right)-\log _{10}\left(18\right)$$
$$\displaystyle \log _{10}\left(\frac{1800}{18}\right)$$
Divide.
$$\displaystyle 1800\div18=100$$
$$\log _{10}\left(100\right)$$
$$\displaystyle 10^2=10*10=100$$
$$\displaystyle \log _{10}\left(10^2\right)$$
$$\displaystyle \log _{10}\left(10^2\right)=2\log _{10}\left(10\right)$$
$$\displaystyle 2\log _{10}\left(10\right)$$
$$\displaystyle \log _{10}\left(10\right)=1$$
Multiply the numbers from left to right.
$$\displaystyle 2*1=\boxed{2}$$
Therefore the correct answer is 2.