Help with math problem?! best answer in an hour!

1. condense: log4(25) - (5/2)log4(x) - (7/2)log4(y) - (3/2)log4(z+9)

*all the numbers that come before parenthesis are the base of the log

2. solve for x: ax + 3 = 2y - 5x

3. 4cotx = cotxsinx

log4(25) - (5/2)log4(x) - (7/2)log4(y) - (3/2)log4(z+9) =

= log4 5^2 - (log4 x^(5/2) + log4 y^(7/2) + log4 (z+9)^(3/2) ) =

= log4 25 - log4(x^(5/2) * y^(7/2) * (z + 9)^3/2)] =

= log4 25 - log4 (x^5 y^7 (z+9)^3)^(1/2)

= 2log4 5 - (1/2) log4 (x^5 y^7 (z+9)^3)

OR

log4 25/(x^5 y^7 (z+9)^3)^(1/2)

2.
ax + 3 = 2y - 5x

ax + 5x = 2y - 3

x(a + 5) = 2y - 3

x = (2y - 3)/(a + 5)

3.

4 (cot x) - (cot x)(sin x) = 0

(cot x)(4 - sin x) = 0

cot x = 0 --> x = pi/2 + kpi (integer k)

4 - sin x = 0 is impossible as sin x is between -1 and +1 for all x

1) log4 [25 / ( (x^(5/2) * y^(7/2) *(z+9)^(3/2) ) ]

2) ax + 3 = 2y - 5x


x(a+5) = 2y - 3 ---> x = (2y-3) / (a+5)


3) 4cotx = cotxsinx

4cotx - cotxsinx = 0

cotx (4 - sinx) = 0

sinx cannot be equal to 4 as -1<= sinx <= 1

cotx = 0

cosx / sinx = cotx = 0 ---> cosx = 0 ---> x = -+ (Pi / 2) + 2kPi where k is an integer, k E Z

-----edit: info on logarithms ---

loga x + loga y = loga xy

loga x - loga y = loga (x/y)

kloga x = loga x^k

loga x = (logb x)/(logb a)

where a is the base and b is any other base of your choosing