What is x if #log_4 x=2 - log_4 (x+6)#?

2 Answers
Jun 8, 2018

Answer:

See process below

Explanation:

In this type of equations, our goal is to arrive to an expresion like
#log_bA=log_bC# from this, we conclude #A=C#

Lets see

#log_4 x=2-log_4(x+6)#

We know that #2=log_4 16#. then

#log_4 x=log_4 16-log_4(x+6)=log_4(16/(x+6))# applying the rule

#log(A/B)=logA-logB#

So, we have #x=16/(x+6)#

#x^2+6x-16=0#
by quadratic formula

#x=(-6+-sqrt(36+64))/2=(-6+-10)/2#

Solutions are #x_1=-8# and #x_2=2# we reject negative and the only valid solution is #x_2#

Jun 8, 2018

Answer:

#x=2#

Explanation:

#"using the "color(blue)"laws of logarithms"#

#•color(white)(x)logx+logy=log(xy)#

#•color(white)(x)log_b x=nhArrx=b^n#

#" add "log_4(x+6)" to both sides"#

#log_4x+log_4(x+6)=2#

#log_4x(x+6)=2#

#x(x+6)=4^2=16#

#x^2+6x-16=0#

#(x+8)(x-2)=0#

#x=-8" or "x=2#

#x>0" and "x+6>0#

#"thus "x=-8" is invalid"#

#rArrx=2" is the solution"#