There are 10 cards. Each card has one number between 1 and 10, so that every number from 1 to 10 appears once.
1 answer:
You might be interested in
First lets start with number six. The only way to solve this is if you determine what "a" and "b" are using the first log they have given to you.
The first variable that I solved for was "a" and }
The same is also true for "b", but when you put both "a" and "b" together the only combination that I have found to work is
Next you plug these numbers in for "a" and "b" on the second equation to get something that looks like this: and the picture below shows where the answer becomes a negative fraction.
https://www.symbolab.com/solver/logarithms-calculator/%20%5Clog_%7B%5Cfrac%7B1%7D%7B2%7D%5Ccdot%5Cfrac%7B1%7D%7B4%7D%7D%5Cleft(%5Cfrac%7B%5Csqrt%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D%7B%5Csqrt%5B3%5D%7B%5Cfrac%7B1%7D%7B4%7D%7D%7D%5Cright)
If you paste that link in your search bar it will give you a even more in depth understanding of how to get this answer
Next is #7, the easier of the two.There are two ways to solve for your answers. According to the graph of this equation there are four possible real solutions. . (This does not account for any complex solutions)
Notice that the bases are conjugates which is why the answers are so "nice"
The key is in the exponents
if then the sum on the conjugates will be 10 so
so
Now for the other two
the solution is also true if
so
the four real solutions are
