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

##### 2 Answers

Jul 6, 2017

See below.

#### Explanation:

Jul 6, 2017

#### 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 #