Solve the equation #(logx)^2-4logx=0#?

2 Answers
Jul 6, 2017

Answer:

See below.

Explanation:

#(logx)^2-4logx=0# or

#(logx-4)logx=0# so we have the solutions

#{(logx-4=0 -> x= e^4),(logx=0->x=1):}#

Jul 6, 2017

Answer:

#x=1# or #x=e^4#

Explanation:

We have:

# (lnx)^2 = ln(x^4) #

Using the rules of logarithms, this can be written as:

# (lnx)^2 = 4lnx #

# :. (lnx)^2 - 4lnx = 0 #
# :. lnx(lnx-4) = 0 #

This would lead to two possible solutions:

Either:

# ln x = 0 => x=e^0 #
# " " => x=1#

Or:

# lnx-4 =0 => ln x =4 #
# " " => x=e^4 #