Thread: Solve for x

Solve for x:

[log_3_(x)]^2 - 4log_9_(x) - 3 = 0


just for reference,
log_3_ means log base 3...


thanks

Originally Posted by Mr_Green

Solve for x:

[log_3_(x)]^2 - 4log_9_(x) - 3 = 0


just for reference,
log_3_ means log base 3...


thanks

Hey Mr Green. just wanted to make sure it was [log_3_(x)]^2 and not log_3_(x)^2. as in its the whole log squared and not just the x

thanks for making sure, but the way i posted it is correct. the whole log is squared, not only the x. so (log base 3 of x)Squared

thanks.

Originally Posted by Jhevon

Hey Mr Green. just wanted to make sure it was [log_3_(x)]^2 and not log_3_(x)^2. as in its the whole log squared and not just the x

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