How do you solve #log_2 x = log_4 (x+6)#?

3 Answers

Using the change of base rule we have that

#logx/log2=log(x+6)/(log2^2) => 2*logx=log(x+6)=> x^2=x+6=> x^2-x-6=0=> (x+2)(x-3)=0=> x_1=3,x_2=-2#

Only #x=3# is acceptable because #x>0#

Nov 14, 2015

We have to use the properties of logarithms.

Explanation:

#log _2x=log_4 (x+6)#
#=> ln x/ln 2= ln (x+6)/ln 4#
#=>ln x/ln 2 = ln (x+6)/ln(2^2)#
#=>ln x/cancel(ln2) = ln (x+6)/(2* cancel(ln2))#
#=>2 ln x= ln(x+6)#
#=> ln (x^2) = ln(x+6)#
#=> x^2=x+6#
#=> x^2-x-6=0#

which is thus reduced to a quadratic equation which can be solved by the quadratic formula to get the two roots: #x=-2 and x=3#

For #x=3#, the arguments within the logarithms in the original equation come out to be positive, thus #x=3# is a valid solution.

But for #x=-2#, one of the logarithmic argument is #-2# , giving rise to #log_2(-2)# which is completely meaningless. Thus #x=-2# is discarded and cannot be considered as a solution.

Nov 14, 2015

Make both sides have a base of #4#.

Explanation:

Change to #4^(log_2(x))=4^(log_4(x+6)#.

Notice that on the right side, the #4# and #log_4# will cancel, leaving just #(x+6)#.
#4^(log_2(x))=x+6#

Change the #4# to #2^2#.
#(2^2)^(log_2(x))=x+6#
#2^(2log_2(x))=x+6#

Remember this logarithm rule.
#2^(log_2(x^2))=x+6#

Like before, the #2# and #log_2# will cancel.
#x^2=x+6#
#x^2-x-6=0#
#(x-3)(x+2)=0#
#x=3 or cancel(x=-2#

Remember that in #log_a(b),b>0#.