Solve for x:
[log_3_(x)]^2 - 4log_9_(x) - 3 = 0
just for reference,
log_3_ means log base 3...
thanks
Note that for some variable a:
log_9_a = [sqrt(log_3_sqrt(a))]^2 = log_3_sqrt(3)
In other words, if we have the log to some base of a number, it is equivalent to the log to the sqrt of the base of the squareroot of the number. How does this work? Here goes.
[log_3_(x)]^2 - 4log_9_(x) - 3 = 0
=> [log_3_(x)]^2 - log_9_(x)^4 - 3 = 0
=> [log_3_(x)]^2 - log_3_(x)^2 - 3 = 0
=> [log_3_(x)]^2 - 2log_3_(x) - 3 = 0
Let y be log_3_x
=> y^2 - 2y - 3 = 0
=> (y - 3)(y + 1) = 0
=> y = 3 or y = -1
so log_3_x = 3 or log_3_x = -1
=> x = 27 or x = 1/3