What is x if #log_4(8x ) - 2 = log_4 (x-1)#?

1 Answer
KillerBunny
Nov 1, 2015

#x=2#

Explanation:

We would like to have an expression like

#log_4(a)=log_4(b)#, because if we had it, we could finish easily, observing that the equation would the solved if and only if #a=b#. So, let's do some manipulations:

  1. First of all, note that #4^2=16#, so #2=log_4(16)#.

The equation then rewrites as

#log_4(8x)-log_4(16)=log_4(x-1)#

But we're still not happy, because we have the difference of two logarithms in the left member, and we want a unique one. So we use

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

So, the equation becomes

#log_4(8x/16)=log_4(x-1)#

Which is of course

#log_4(x/2)=log_4(x-1)#

Now we are in the desired form: since the logarithm is injective, if #log_4(a)=log_4(b)#, then necessarily #a=b#. In our case,

#log_4(x/2)=log_4(x-1) iff x/2 = x-1#

Which is easily solve into #x=2x-2#, which yields #x=2#

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