I don’t know how the solution works with this one.
A achi_kun New member Joined Oct 20, 2020 Messages 7 Oct 28, 2020 #1 I don’t know how the solution works with this one.
D Deleted member 4993 Guest Oct 28, 2020 #2 achi_kun said: View attachment 22669 I don’t know how the solution works with this one. Click to expand... If I were to do this problem, I would use the fact: logn(n) = 1 Please show us what you have tried and exactly where you are stuck. Please follow the rules of posting in this forum, as enunciated at: READ BEFORE POSTING Please share your work/thoughts about this problem.
achi_kun said: View attachment 22669 I don’t know how the solution works with this one. Click to expand... If I were to do this problem, I would use the fact: logn(n) = 1 Please show us what you have tried and exactly where you are stuck. Please follow the rules of posting in this forum, as enunciated at: READ BEFORE POSTING Please share your work/thoughts about this problem.
pka Elite Member Joined Jan 29, 2005 Messages 11,976 Oct 28, 2020 #3 If I were to do this I would rewrite it as \(\dfrac{\log(3x-1)}{\log(7)}>1\). Recall that the logarithm is one-to-one.
If I were to do this I would rewrite it as \(\dfrac{\log(3x-1)}{\log(7)}>1\). Recall that the logarithm is one-to-one.
skeeter Elite Member Joined Dec 15, 2005 Messages 3,216 Oct 28, 2020 #4 [MATH]\log_7(7) = 1 \implies \log_7(3x-1) > 1 \text{ if ... ?}[/MATH]