Answer:
(a)![\log_3(\dfrac{81}{3})=3](https://tex.z-dn.net/?f=%5Clog_3%28%5Cdfrac%7B81%7D%7B3%7D%29%3D3)
(b)![\log_5(\dfrac{625}{25})=2](https://tex.z-dn.net/?f=%5Clog_5%28%5Cdfrac%7B625%7D%7B25%7D%29%3D2)
(c)![\log_2(\dfrac{64}{8})=3](https://tex.z-dn.net/?f=%5Clog_2%28%5Cdfrac%7B64%7D%7B8%7D%29%3D3)
(d)![\log_4(\dfrac{64}{16})=1](https://tex.z-dn.net/?f=%5Clog_4%28%5Cdfrac%7B64%7D%7B16%7D%29%3D1)
(e)![\log_6(36^4)=8](https://tex.z-dn.net/?f=%5Clog_6%2836%5E4%29%3D8)
(f)![\log(100^3)=6](https://tex.z-dn.net/?f=%5Clog%28100%5E3%29%3D6)
Step-by-step explanation:
Let as consider the given equations are
.
(a)
![\log_3(\dfrac{81}{3})=\log_3(27)](https://tex.z-dn.net/?f=%5Clog_3%28%5Cdfrac%7B81%7D%7B3%7D%29%3D%5Clog_3%2827%29)
![\log_3(\dfrac{81}{3})=\log_3(3^3)](https://tex.z-dn.net/?f=%5Clog_3%28%5Cdfrac%7B81%7D%7B3%7D%29%3D%5Clog_3%283%5E3%29)
![[\because \log_aa^x=x]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%5Clog_aa%5Ex%3Dx%5D)
(b)
![[\because \log_aa^x=x]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%5Clog_aa%5Ex%3Dx%5D)
(c)
![\log_2(\dfrac{64}{8})=\log_2(8)](https://tex.z-dn.net/?f=%5Clog_2%28%5Cdfrac%7B64%7D%7B8%7D%29%3D%5Clog_2%288%29)
![\log_2(\dfrac{64}{8})=\log_2(2^3)](https://tex.z-dn.net/?f=%5Clog_2%28%5Cdfrac%7B64%7D%7B8%7D%29%3D%5Clog_2%282%5E3%29)
![[\because \log_aa^x=x]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%5Clog_aa%5Ex%3Dx%5D)
(d)
![\log_4(\dfrac{64}{16})=\log_4(4)](https://tex.z-dn.net/?f=%5Clog_4%28%5Cdfrac%7B64%7D%7B16%7D%29%3D%5Clog_4%284%29)
![[\because \log_aa^x=x]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%5Clog_aa%5Ex%3Dx%5D)
(e)
![\log_6(36^4)=\log_6((6^2)^4)](https://tex.z-dn.net/?f=%5Clog_6%2836%5E4%29%3D%5Clog_6%28%286%5E2%29%5E4%29)
![\log_6(36^4)=\log_6(6^8)](https://tex.z-dn.net/?f=%5Clog_6%2836%5E4%29%3D%5Clog_6%286%5E8%29)
![[\because \log_aa^x=x]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%5Clog_aa%5Ex%3Dx%5D)
(f)
![\log(100^3)=\log((10^2)^3)](https://tex.z-dn.net/?f=%5Clog%28100%5E3%29%3D%5Clog%28%2810%5E2%29%5E3%29)
![\log(100^3)=\log(10^6)](https://tex.z-dn.net/?f=%5Clog%28100%5E3%29%3D%5Clog%2810%5E6%29)
![[\because \log10^x=x]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%5Clog10%5Ex%3Dx%5D)
Answer:
Ф = ![\frac{\pi }{4} ,\frac{5\pi}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpi%20%7D%7B4%7D%20%2C%5Cfrac%7B5%5Cpi%7D%7B4%7D)
Step-by-step explanation:
It is a bit difficult to input the work here, so I uploaded an image
- First we can use the trig identities to change sec²(Ф) to tan²(Ф) + 1
- Then we can combine like terms
- Then we can factor this as a polynomial function
- Then we can set each term equal to zero and solve for Ф
- The first term tan(Ф) - 2 = 0 has no solution because tan(Ф) ≠ -2 anywhere
- The second term tan(Ф) - 1 = 0 has two solutions of
and
so these are the solutions to the problem
Answer:
4 bouquets
Step-by-step explanation:
5/5 = 20 flower bouquets.
Find out how much flower bouquets there are when it's 1/5.
5/5 = 20
1/5 = 20 ÷ 5 = 4
Since tulips are 1/5 as 5/5 - 4/5 = 1/5, 4 bouquets are tulips.
Answer:CD = x = 4 units
Explanation:We are given that the two triangles are similar. This means that we can set the similarity proportionality as follows:
![\frac{BC}{CD} = \frac{CA}{CE} = \frac{AB}{ED}](https://tex.z-dn.net/?f=%20%5Cfrac%7BBC%7D%7BCD%7D%20%3D%20%20%5Cfrac%7BCA%7D%7BCE%7D%20%3D%20%20%5Cfrac%7BAB%7D%7BED%7D%20%20%20)
W e have:
BC = 20 - x
AC = 20
CD = x
CE = 5
Substitute in the above proportionality and solve for x as follows:
![\frac{20-x}{x} = \frac{20}{5} = 4](https://tex.z-dn.net/?f=%20%5Cfrac%7B20-x%7D%7Bx%7D%20%3D%20%20%5Cfrac%7B20%7D%7B5%7D%20%3D%204)
4x = 20 - x
5x = 20
x = 4
Based on he above:
CD = 4 units
BC = 20 - x = 20 - 4 = 16 units
Hope this helps :)