Solve for x:

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

just for reference,

log_3_ means log base 3...

thanks

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- Mar 4th 2007, 10:18 AMMr_GreenSolve for x
Solve for x:

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

just for reference,

log_3_ means log base 3...

thanks - Mar 4th 2007, 10:31 AMJhevon
- Mar 4th 2007, 10:39 AMMr_Green
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. - Mar 4th 2007, 11:03 AMJhevon
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