What is x if #3ln2+ln(x^2)+2=4#?

2 Answers
Nov 8, 2015

#x=e^{1-3/2 ln(2)}#

Explanation:

Isolate the term involving #x#:

#ln(x^2) = 4-2-3ln(2) = 2-3ln(2)#

Use the property of logarithm #ln(a^b)=bln(a)#:

#2ln(x)=2-3ln(2)#

Isolate the term involving #x# again:

#ln(x)=1-3/2 ln(2)#

Take the exponential of both terms:

#e^{ln(x)} = e^{1-3/2 ln(2)}#

Consider the fact that exponential and logarithm are inverse functions, and thus #e^{ln(x)} =x#

#x=e^{1-3/2 ln(2)}#

Nov 8, 2015

#x=+-(esqrt2)/4#

Explanation:

#[1]" "3ln2+ln(x^2)+2=4#

Subtract #2# from both sides.

#[2]" "3ln2+ln(x^2)+2-2=4-2#

#[3]" "3ln2+ln(x^2)=2#

Property: #alog_bm=log_bm^a#

#[4]" "ln2^3+ln(x^2)=2#

#[5]" "ln8+ln(x^2)=2#

Property: #log_bm+log_bn=log_b(mn)#

#[6]" "ln(8x^2)=2#

#[7]" "log_e(8x^2)=2#

Convert to exponential form.

#[8]" "hArre^2=8x^2#

Divide both sides by #8#.

#[9]" "e^2/8=x^2#

Subtract #e^2/8# from both sides.

#[10]" "x^2-e^2/8=0#

Difference of two squares.

#[11]" "(x+sqrt(e^2/8))(x-sqrt(e^2/8))=0#

#[12]" "(x+e/(2sqrt2))(x-e/(2sqrt2))=0#

Rationalize.

#[13]" "(x+(esqrt2)/4)(x-(esqrt2)/4)=0#

Therefore: #color(blue)(x=+-(esqrt2)/4)#