What are possible values of x if # 2logx

Question

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

Genesis Allen · Accepted Answer

No possible solutions.

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.

Hazel Anglin · Answer

undefined The possible values of $ x $ are $ x > 1 $ and $ x \neq 1 $.

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

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