Using logarithmic laws: log(a) + log(b) = log(ab)
log₄ (x + 9) + log₄ (x + 21) = log₄ [(x + 9)(x + 21)] = 3
Since we want the equation in terms of x, the next intuitive step is take 4 to the power of both sides, because the left hand side will simplify easily.
![4^{log_4 (x + 9)(x + 21)} = 4^{3}](https://tex.z-dn.net/?f=4%5E%7Blog_4%20%28x%20%2B%209%29%28x%20%2B%2021%29%7D%20%3D%204%5E%7B3%7D)
![(x + 9)(x + 21) = 64](https://tex.z-dn.net/?f=%28x%20%2B%209%29%28x%20%2B%2021%29%20%3D%2064)
![x^{2} + 9x + 21x + 189 = 64](https://tex.z-dn.net/?f=x%5E%7B2%7D%20%2B%209x%20%2B%2021x%20%2B%20189%20%3D%2064)
Solving the quadratic we get:
![x^{2} + 30x + 125 = 0](https://tex.z-dn.net/?f=x%5E%7B2%7D%20%2B%2030x%20%2B%20125%20%3D%200)
![(x + 5)(x + 25) = 0](https://tex.z-dn.net/?f=%28x%20%2B%205%29%28x%20%2B%2025%29%20%3D%200)
![x = -5, -25](https://tex.z-dn.net/?f=x%20%3D%20-5%2C%20-25)
BUT, for the logarithmic equation to be defined, x > -9, so x≠ -25
Thus, the only solution is x = -5.
Answer:
the answer is 0.7>0.68
Step-by-step explanation:
because the seven is bigger than 6.8
Step-by-step explanation:
the answer is in the picture above
Step-by-step explanation:
![y = 2x - 7...(1) \\ y = 5x + 4...(2) \\ from \: equations \: (1) \: and \: (2) \\ 5x + 4 = 2x - 7 \\ \therefore \: 5x - 2x = - 7 - 4 \\ \therefore \: 3x = - 11\\ \huge \red{ \boxed{\therefore \: x = - \frac{11}{3} }}\\ substitutig \: x = - \frac{11}{3} in \: equation \: (1) \\ y = 2 \times (- \frac{11}{3}) - 7 \\ \\ \therefore \: y = - \frac{22}{3} - 7 \\ \\ \therefore \: y = \frac{ - 22 - 21}{3} \\ \\ \huge \orange{ \boxed{\therefore \: y = - \frac{ 43}{3}}}\\\\(x, \:y) = \{- \frac{11}{3}, - \frac{ 43}{3}\} \:](https://tex.z-dn.net/?f=y%20%3D%202x%20-%207...%281%29%20%5C%5C%20y%20%3D%205x%20%2B%204...%282%29%20%5C%5C%20from%20%5C%3A%20equations%20%5C%3A%20%281%29%20%5C%3A%20and%20%5C%3A%20%282%29%20%5C%5C%205x%20%2B%204%20%3D%202x%20-%207%20%5C%5C%20%20%20%5Ctherefore%20%5C%3A%205x%20-%202x%20%3D%20%20-%207%20-%204%20%5C%5C%20%20%5Ctherefore%20%5C%3A%203x%20%3D%20%20-%2011%5C%5C%20%20%20%5Chuge%20%5Cred%7B%20%5Cboxed%7B%5Ctherefore%20%5C%3A%20x%20%3D%20%20-%20%20%20%5Cfrac%7B11%7D%7B3%7D%20%7D%7D%5C%5C%20%20substitutig%20%5C%3A%20x%20%3D%20%20-%20%20%20%5Cfrac%7B11%7D%7B3%7D%20in%20%5C%3A%20equation%20%5C%3A%20%281%29%20%5C%5C%20y%20%3D%202%20%5Ctimes%20%28-%20%20%20%5Cfrac%7B11%7D%7B3%7D%29%20-%207%20%5C%5C%20%20%5C%5C%20%5Ctherefore%20%5C%3A%20y%20%3D%20%20-%20%20%5Cfrac%7B22%7D%7B3%7D%20%20-%207%20%5C%5C%20%20%5C%5C%20%20%5Ctherefore%20%5C%3A%20y%20%3D%20%20%20%5Cfrac%7B%20-%2022%20-%2021%7D%7B3%7D%20%5C%5C%20%20%5C%5C%20%20%5Chuge%20%5Corange%7B%20%5Cboxed%7B%5Ctherefore%20%5C%3A%20y%20%3D%20%20%20-%20%5Cfrac%7B%2043%7D%7B3%7D%7D%7D%5C%5C%5C%5C%3C%2Fp%3E%3Cp%3E%28x%2C%20%5C%3Ay%29%20%3D%20%5C%7B-%20%20%20%5Cfrac%7B11%7D%7B3%7D%2C%20-%20%5Cfrac%7B%2043%7D%7B3%7D%5C%7D%20%5C%3A%20)