log ,base, 3, 3 = 1
(there is only 1 required to make 1 as log ,base 3, 3 = 1)
log ,base 9, 3+ log, base 9, 3 = 1
(there is only 2 required to make 1 as log ,base 9, 3 = 1/2)
log ,base 27, 3 + ⋯ = 1
(3 would be required here to make 1 as log ,base 27, 3 = 1/3)
How many log ,base81, 3 do you need to add together to make one?
4 because log ,base 81, 3 is equal to 1/4. (3 to the power of 4 is 81. So to go to 3 from 81 you put 81 to the power of 1/4)
Can we choose integers x and y so that:
log6x+log6y=1?
How many different ways are there to do this?
There are as many solutions as there are factors of 6. As long as the integers multiply together to get 6 they will add up to 1 when logged to base 6. In this case 6x1=6 and 3x2=6 so there are 2 integers that will work. Negatives will not work because a negative cannot be logged. log,base a,b =c is the same as a to the power of c= b and no power will give a negative answer unless a is negative.
How about using log12?
Same explanation as above: positive factors of 12. The integer answers are: 12 & 1, 6 & 2, 3 & 4
How about using log24?
Same explanation as above: positive factors of 12. The integer answers are:
24 & 1, 12 & 2, 6 & 4, 3 & 8