Question #82e26

3 Answers
May 30, 2016

Answer:

#x = -4# and #x = 1/2#

Explanation:

"A" is rather similar to "4" so I suppose there is a typo in between.
Putting #4# instead "A" it works. So

#log_2x+log_2(2x+7)=log_2 2^2 = 2#

#log_2(x(2x+7))=log_2 4 equiv x(2x+7)=4#

Solving for #x#

#x = -4# and #x = 1/2#

May 30, 2016

Answer:

Here we go and obtain
#x=1/2#

Explanation:

#log_2x+log_2(2x+7)=log_2A# .......(1)
Solve:
#log_2x+log_2(2x+7)=2# .......(2)

We see from inspection that LHS of both equations (1) and (2) are equal. Therefore, RHS of both the equations must be equal.
#=> log_2A=2#
Writing this in exponential form by definition of #log# we get
#2^2=A#,
or #A=4#
Inserting the value of #A# in (1), equation (2) becomes
#log_2x+log_2(2x+7)=log_2 4#
LHS can be written using the rule for, addition of #log# functions as
#log_2x+log_2(2x+7)=log_2 [x(2x+7)]#
or #log_2 (2x^2+7x)#
Equating with RHS
#log_2 (2x^2+7x)=log_2 4#
Taking #"anti" log_2# of both sides, we get
#2x^2+7x=4#
or #2x^2+7x-4=0#,

Using the split the middle term method
#2x^2+8x-x-4=0#
or #2x(x+4)-(x+4)=0#
or #(x+4)(2x-1)=0#
We get two roots by putting each factor #=0#

  1. #x+4=0#
    or #x=-4#
  2. #2x-1=0#
    or #x=1/2#

Since #log# of a #-ve# number is not defined, therefore, only valid solution is #x=1/2#

May 30, 2016

Answer:

A=4 and #x = 1/2#

Explanation:

The derivations for x = -4 and 1/2 are as in other two answers.

#x-4# is inadmissible.

log of negative argument is unreal.

#log_2 (-4)=log_2(4i^2)#

#=log_2 4+log_2(i^2)#

#=.log_2(2^2)+2 log_2(i^2)#

#=2(log_2 2+log_2 i)#

The RHS #log_2 A# is a hint to solve for the RHS in the second equation. Of course, the second is sufficient, without the first..