Math Sucks..will you please help me i am struggling to hold on to the C that i have in this class???

2log(subscript3) y=log(subscript3) 4+log (subscript3) (y+8)








solve please! AHH

Math Sucks..will you please help me i am struggling to hold on to the C that i have in this class???
use properties of logs...


r log a = log (a^r)


log a + lob b = log (ab)





for now, all logs are base 3:


2log y = log 4 + log (y +8)





use the two properties above to make this:


log (y^2) = log [4(y + 8)]


log (y^2) = log (4y + 32)





you have two logs with a common base 3, so the arguments must be equal.. drop the logs and solve the equation


y^2 = 4y + 32


y^2 - 4y - 32 = 0


(y + 4)(y - 8) = 0


y +4 = 0 or y - 8 = 0


y = -4 or y = 8





but.... the domain of log x is x%26gt;0...


so discard -4 because log (-4) is undefined





solution: y = 8





check: 2 log 8 = log 4 + log (8 + 8)


log(64) = log 4 + log 16


log (64) = log (64) check!
Reply:Hi!





Note: log[a] (x) means log (subscript a) (x).


In order to solve this problem, recall the following logarithm rules:


log[a] (x) + log[a] (y) = log[a] (xy)


log[a] (x) - log[a] (y) = log[a] (x/y)


b*log[a] (x) = log[a] (x^b)


log[a] (x) = log[a] (y) implies x = y.


[Please note that inorder to apply these rules, the base (i.e. a) of the logarithm must be the same throughout the problem.]





We want to solve:


2 log[3] (y) = log[3] (4) + log[3] (y+8)


log[3] (y^2) = log[3] (4) + log[3] (y+8)


log[3] (y^2) - log[3] (y+8) = log[3] (4)


log[3] ((y^2)/(y+8)) = log[3] (4)


(y^2)/(y+8) = 4


(y^2) = 4*(y+8)


y^2 - 4y - 32 = 0


Factorising...


(y - 8) (y + 4) = 0


So, y must be either 8 or -4.


Note that the solution of y must satisfy:


2 log[3] (y) = log[3] (4) + log[3] (y+8)


But, we can't find logarithms for negative numbers.


Hence, the answer must be: y = 8.





Hope this helps.
Reply:use properties of logs...


r log a = log (a^r)


log a + lob b = log (ab)





for now, all logs are base 3:


2log y = log 4 + log (y +8)





use the two properties above to make this:


log (y^2) = log [4(y + 8)]


log (y^2) = log (4y + 32)





you have two logs with a common base 3, so the arguments must be equal.. drop the logs and solve the equation


y^2 = 4y + 32


y^2 - 4y - 32 = 0


(y + 4)(y - 8) = 0


y +4 = 0 or y - 8 = 0


y = -4 or y = 8





but.... the domain of log x is x%26gt;0...


so discard -4 because log (-4) is undefined





solution: y = 8





check: 2 log 8 = log 4 + log (8 + 8)


log(64) = log 4 + log 16


log (64) = log (64) check!








do u get it now???
Reply:so raise both sides to the power of 3 to get rid of the log (sub 3):





3^(2log(subscript 3)y) =


3^(log(subscript3)4+log(subscript3)(y+...





now use log rules on the LHS to put the 2 as y^2





3^(log(sub3)y^2) = 3^(log(sub3)4+log(sub3)(y+8))





now 3^(log(sub3)y^2) is just y^2 (more log rules), so:





y^2 = 3^(log(sub3)4+log(sub3)(y+8))





now, x^(a+b) = x^a*x^b, so the RHS becomes:





y^2 = 3^(log(sub3)4) * 3^(log(sub3)(y+8))





now using the fact that 3^(log(sub3)4) = 4 and 3^(log(sub3)(y+8) = y+8 gives us:





y^2 = 4*(y+8)





y^2 = 4y+32





y^2-4y-32=0





(y+4)(y-8) = 0





so y = -4 or y =8