How do you solve #log(x + 10) - log(x) = 2log(5)#?

1 Answer
Jacob F.
Sep 11, 2015

#x = 5/12#

Explanation:

#log(x + 10) - logx = 2log5#

Our first step is to rewrite the equation using some laws of logarithms, specifically, #logA - logB = log (A/B)#. This law allows us to rewrite the left-hand side of the equation as

#log((x + 10)/x) = 2log5#.

Another law of logarithms, #A log B = log B^A#, allows us to rewrite the right-hand side equivalently as

#log((x + 10)/x) = log5^2#
#= log 25#

Now, you didn't specify a base for the #log# function here, so I will assume that #log# means base-2 logarithm. Still, whether the base is 2, 10, e, or whatever, it actually doesn't matter... the answer will be the same. You'll see why in a moment.

Raise 2 to both sides:

#2^log((x + 10)/x) = 2^log25#

The exponential and the logarithm are inverse functions, so the base-2 and the logarithms will cancel:

#(x + 10)/x = 25#

From here, we just need to use some simple algebra, multiplying both sides by #x#:

#x + 10 = 25x#

and then subtracting #x# from both sides:

#10 = 24x#

And then simplify to arrive at our final answer #x#:

#x = 5/12#

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