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.