Question 1134465
looks like you are solving two equations simultaneouosly with the twist that they involve logs.


the two equations are:


log27(x) + log4(y) = 5
log27(x) - log4(y) = 1


subtract the second equation from the first to get:


2 * log4(y) = 4


divide both sides of this equation by 2 to get log4(y) = 2


since log4(y) = 2, then both equations become:


log27(x) + 2 = 5
log27(x) - 2 = 1


add these two equations together to get:


2 * log27(x) = 6


divide this equaion by 2 to get log27(x) = 3.


your solution so far is:


log27(x) + log4(y) = 5 which becomes 3 + 2 = 5
log27(x) - log4(y) = 1 which becomes 3 - 2 = 1


the equations have been solved simultaneously and what is left is to find the value of x and y.


it was already determined that log27(x) = 3 and log4(y) = 2.


log27(x) = 3 if and only if 27^3 = x.
this makes x = 19683.


log4(y) = 2 if and only if 4^2 = y.
this makes y = 16.


your solution is that x = 19683 and y = 16.


you can confirm by using the base log conversion formula and your calculator.


log27(19683) = log(19683) / log(27) = 3


log4(16) = log(16) / log(4) = 2


solution is confirmed to be good.