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Logarithm problem
- Thread starter Nazariy
- Start date
if it wasn't for that 1/3 in the end, I could just compare the powers of the two, because the base is two in both cases. So perhaps think of a number with the base 2 that when raised to power x gives 1/3 ?
hmm, dont think that is gonna help. it would be two to the power of -1.58
D
Deleted member 4993
Guest
I need some help with this...
View attachment 4998
Here are my workings.. All I have done it appears is came back to "simplified" original. It appears there are two solutions here actually.
View attachment 4999
22x+1 = 3*2x -1
22x * 2= 3*2x -1
1 = 2x
x = 0
22x+1 = 3*2x -1
22x * 2= 3*2x -1
1 = 2x
x = 0
x=-1 as well..
D
Deleted member 4993
Guest
x=-1 as well..
1 = 2-1 = 1/2 ??????!!!!!!! .......... certainly not
22x+1 = 3*2x -1
22x * 2= 3*2x -1
1 = 2x
x = 0
And I don't follow what happened from here: 22x * 2= 3*2x -1
To here: 1 = 2x
1 = 2-1 = 1/2 ??????!!!!!!! .......... certainly not
2^(2*(-1)+1) = 0.5
3*(2^(-1))-1 = 0.5
....
D
Deleted member 4993
Guest
And I don't follow what happened from here: 22x * 2= 3*2x -1
I skipped a few steps ... I thought you could fill it in...
To here: 1 = 2x
22x * 2= 3*2x -1
substitute u = 2x
then we have:
2u2 - 3u + 1 = 0
(u - 1)(2u - 1) = 0
so
u - 1 = 0 → 2x - 1 = 0 → x = 0
or
2u - 1 = 0 → ?????
22x * 2= 3*2x -1
substitute u = 2x
then we have:
2u2 - 3u + 1 = 0
(u - 1)(2u - 1) = 0
so
u - 1 = 0 → 2x - 1 = 0 → x = 0
or
2u - 1 = 0 → ?????
Sorry this problem was a bit irritating.. Couldn't figure out the steps you skipped.
Brilliant, now we have -1 as well. Thank you for your help.