What are possible values of x if # 2logx<log(2x-1)?#?

1 Answer
Nov 9, 2015

Answer:

No possible solutions.

Explanation:

First, it is always a good idea to identify the domain of your logarithm expressions.

For #log x#: the domain is #x > 0#
For #log(2x-1)#: the domain is # 2x - 1 > 0 <=> x > 1/2#

This means that we only need to consider #x# values where #x > 1/2# (the intersection of the two domains) since otherwise, at least one of the two logarithm expressions is not defined.

Next step: use the logarithm rule #log(a^b) = b * log(a)# and transform the left expression:
# 2 log(x) = log(x^2)#

Now, I'm assuming that the basis of your logarithms is #e# or #10# or a different basis #>1#. (Otherwise, the solution would be quite different).

If this is the case, #log(f(x)) < log(g(x)) <=> f(x) < g(x)# holds.

In your case:
#log(x^2) < log(2x - 1)#
#<=> x^2 < 2x - 1#
#<=> x^2 - 2 x + 1 < 0#
#<=> (x-1)^2 < 0#

Now, this is a false statement for all real numbers #x# since a quadratic expression is always #>=0#.

This means that (under the assumption that your logarithm basis is indeed #>1#) your inequality has no solutions.