Given:


To find:
The value of
.
Solution:
We have,


Using properties of log, we get
![\left[\because \log_a\dfrac{m}{n}=\log_am-\log_an\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbecause%20%5Clog_a%5Cdfrac%7Bm%7D%7Bn%7D%3D%5Clog_am-%5Clog_an%5Cright%5D)
![[\log x^n=n\log x]](https://tex.z-dn.net/?f=%5B%5Clog%20x%5En%3Dn%5Clog%20x%5D)
Substitute
and
.



Therefore, the value of
is
.
Answer:
The answer is 5.
Step-by-step explanation:
Yes I think it really is the correct answer
Answer: 2 meters.
Step-by-step explanation:
Let w = width of the cement path.
Dimensions of pool : Length = 15 meters , width = 9 meters
Area of pool = length x width = 15 x 9 = 135 square meters
Along width cement path, the length of region = 
width = 
Area of road with pool = 

Area of road = (Area of road with pool ) -(area of pool)
![\Rightarrow\ 112 =4w^2+48w+135- 135\\\\\Rightarrow\ 112= 4w^2+48w\\\\\Rightarrow\ 4 w^2+48w-112=0\\\\\Rightarrow\ w^2+12w-28=0\ \ \ [\text{Divide both sides by 4}]\\\\\Rightarrow\ w^2+14w-2w-28=0\\\\\Rightarrow\ w(w+14)-2(w+14)=0\\\\\Rightarrow\ (w+14)(w-2)=0\\\\\Rightarrow\ w=-14\ or \ w=2](https://tex.z-dn.net/?f=%5CRightarrow%5C%20112%20%3D4w%5E2%2B48w%2B135-%20135%5C%5C%5C%5C%5CRightarrow%5C%20112%3D%204w%5E2%2B48w%5C%5C%5C%5C%5CRightarrow%5C%204%20w%5E2%2B48w-112%3D0%5C%5C%5C%5C%5CRightarrow%5C%20w%5E2%2B12w-28%3D0%5C%20%5C%20%5C%20%5B%5Ctext%7BDivide%20both%20sides%20by%204%7D%5D%5C%5C%5C%5C%5CRightarrow%5C%20w%5E2%2B14w-2w-28%3D0%5C%5C%5C%5C%5CRightarrow%5C%20w%28w%2B14%29-2%28w%2B14%29%3D0%5C%5C%5C%5C%5CRightarrow%5C%20%28w%2B14%29%28w-2%29%3D0%5C%5C%5C%5C%5CRightarrow%5C%20%20w%3D-14%5C%20or%20%5C%20w%3D2)
width cannot be negative, so w=2 meters
Hence, the width of the road = 2 meters.