I'm going to use the notation log(2,x) to indicate "log base 2 of x". The first number is the base while the second is the expression inside the log (aka the argument of the log)
log(2,x) + log(2,(x-6)) = 4 log(2,x*(x-6)) = 4 x*(x-6) = 2^4 x*(x-6) = 16 x^2-6x = 16 x^2-6x-16 = 0 (x-8)(x+2) = 0 x-8 = 0 or x+2 = 0 x = 8 or x = -2
Recall that the domain of log(x) is x > 0. So x = -2 is not allowed. The same applies to log(2,x) as well.
Only x = 8 is a proper solution.
------------------------
You can use the change of base rule to check your work log base 2 of x = log(2,x) = log(x)/log(2) log(2,(x-6)) = log(x-6)/log(2)
Sin θ=sqrt(1-<span>cos^2 θ) </span>sin θ=sqrt[1-(5/13)^2] sin θ=sqrt[1-(5)^2/(13)^2] sin θ=sqrt[1-25/169] sin θ=sqrt[(169-25)/169] sin θ=sqrt(144/169) sin θ=sqrt(144)/sqrt(169) sin θ=12/13
We can set up an equation to find the answer. 21/x=0.35/1 because we're trying to find how many starbursts make up 100% of the package. So then to solve, we take 21x1 and then divide that by 0.35. From this we find that x=60.
