How do you solve # log x + log (x+1) = 1#?

1 Answer
George C.
Dec 26, 2015

Derive a quadratic, one of whose roots is an acceptable solution of the equation, namely:

#x = (sqrt(41)-1)/2#

Explanation:

From the basic properties of logs we have:

#log 10 = 1 = log x + log(x+1) = log(x(x+1))#

Since #log# is a one-one function from #(0,oo) -> RR#, we require:

#10 = x(x+1)#

That is:

#x^2+x-10 = 0#

This is of the form #ax^2+bx+c# with #a=1#, #b=1# and #c=-10#

Use the quadratic formula to find:

#x = (-b+-sqrt(b^2-4ac))/(2a) = (-1+-sqrt(41))/2#

Since #sqrt(41) > 1# only one of these roots is positive, namely:

#x = (-1+sqrt(41))/2 = (sqrt(41)-1)/2#

We can discard the other root of the quadratic since we are dealing with #log# as a function of non-negative Real numbers, so we disallow #log# of negative numbers.

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