How do you solve #Log_2x + log_2(x+2) = log_2(6x+1)#?

3 Answers
Martin C.
Apr 18, 2018

#x_1=2+sqrt(5) or x_2=2-sqrt(5)#

Explanation:

#Log_2x + log_2(x+2) = log_2(6x+1)#

Remember that #log(a)+log(b)=log(a*b)#

#Log_2(x*(x+2)) = log_2(6x+1)#
#x*(x+2)=6x+1#
#x^2+2x=6x+1|-2x#
#x^2=4x+1|-4x-1#
#x^2-4x-1=0#
#(x-2)^2-1-4=0|+5#
#(x-2)^2=5|sqrt()|+2#
#x=2+-sqrt(5)#
#x_1=2+sqrt(5) or x_2=2-sqrt(5)#
#x_2<0 -> log(x_2)∈CC#

Yahia M.
Apr 18, 2018

#x=2+sqrt5#

Explanation:

#log_2x+log_2(x+2)=log_2(6x+1)#

#color(green)(loga+logb=logab#

#log_2x(x+2)=log_2(6x+1)#

By taking both sides for powers of 2 like this:

#2^(log_2x(x+2))=2^(log_2(6x+1)#

#color(green)(a^(log_ax)=x#

Thus

#x(x+2)=6x+1#

#x^2+2x=6x+1#

#x^2-4x-1=0#

#x=2+sqrt5#

#" OR " #

#x=2-sqrt5##color(red)" refused as it doesn't satisfy the equation"#

Somebody N. · Martin C.
Apr 18, 2018

#color(blue)(x=2+sqrt(5))#

Explanation:

By the laws of logarithms:

#log_a(b)+log_a(c)=log_a(bc)#

#log_a(b)=log_a(c)<=>b=c#

# #

#log_2(x)+log_2(x+2)=log_2(6x+1)#

#log_2(x(x+2))=log_2(6x+1)#

#log_2(x^2+2x)=log_2(6x+1)#

#:.#

#x^2+2x=6x+1#

#x^2-4x-1=0#

By quadratic formula:

#x=(4+-sqrt(16+4))/2=(4+-2sqrt(5))/2=2+-sqrt(5)#

We need to test this:

#x=2-sqrt(5)<0#

For:

#log_2(x)# This is undefined for real mumbers.

So only the soloution:

#color(blue)(x=2+sqrt(5))#

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