H hrm Sep 2017 4 0 london Sep 16, 2017 #1 ok I hate logs so how would I simplify: z= 3- 3 log(x+y) thank you

skeeter MHF Helper Jun 2008 16,217 6,765 North Texas Sep 16, 2017 #2 hrm said: ok I hate logs so how would I simplify: z= 3- 3 log(x+y) Click to expand... one "simplifies" an expression or "solves" an equation. you've posted an equation ... what are you solving for?

hrm said: ok I hate logs so how would I simplify: z= 3- 3 log(x+y) Click to expand... one "simplifies" an expression or "solves" an equation. you've posted an equation ... what are you solving for?

S SlipEternal MHF Helper Nov 2010 3,728 1,571 Sep 16, 2017 #3 $z=3 \big(1-\log (x+y)\big)=3\log\left( \dfrac{e}{x+y}\right) $ Which one do you think is "simpler"? Last edited: Sep 16, 2017

H hrm Sep 2017 4 0 london Sep 16, 2017 #5 sorry guys I posted the question wrong! log z = 3 - 3log(x+y)

skeeter MHF Helper Jun 2008 16,217 6,765 North Texas Sep 16, 2017 #7 hrm said: sorry guys I posted the question wrong! log z = 3 - 3log(x+y) Click to expand... $z = e^{3-3\log(x+y)} = \dfrac{e^3}{e^{\log(x+y)^3}} = \left(\dfrac{e}{x+y}\right)^3$

hrm said: sorry guys I posted the question wrong! log z = 3 - 3log(x+y) Click to expand... $z = e^{3-3\log(x+y)} = \dfrac{e^3}{e^{\log(x+y)^3}} = \left(\dfrac{e}{x+y}\right)^3$