How do you solve #log_4 (x + 4) - log_4 (x - 4) = log_4 3#?

Question

1208 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-06T10:23:12

#x=8#

To solve this kind of equations, the strategy is to manipulate the expressions in order to arrive at something like

#log(X) = log(Y)#

To deduce that #X=Y#, since the logarithm function is injective.

So, the only thing we need to do in this case it to remember that

#log(a)-log(b) = log(a/b)#

So you have that

#log_4((x+4)/(x-4)) = log_4(3)#

Which is true if and only if

#(x+4)/(x-4) = 3#

If #x \ne 4#, we can multiply by #x-4# both terms and get

#x+4 = 3x-12 \to 2x=16 \to x=8#