Question 634748: I am not sure how to determine b:
y=log(base)b X
when it passes through the point (729,6) determine b
please help, thank you
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! 
If the point (729, 6) is on the graph of this equation then those coordinates must fit the equation. Replacing th ex and y in the equation with the coordinates of the point we get:
Now we just solve for b.
We want an equation that says
b = something
To get this we must somehow extract the b from inside the logarithm. To do this we rewrite the equation in exponential form. In general,  is equivalent to  . Using this pattern on our equation we get:
There are several ways to solve this. One way is to find the 6th root of each side. But this can be tricky to get right, especially if the 6th root of 729 is not a whole number. Another way is to use factoring. This takes a little bit of effort but it is safer. This is how we will proceed. First we want one side to be zero. Subtracting 729 from each side:
Now we try to factor. The GCF is 1. Does this fit one of the factoring patterns? We have two patterns with two terms with a "-" between them:
 and
So is  a difference of squares, a difference of cubes or neither? Since  or  , it can be considered both a perfect square and a perfect cube! And 729 is also a perfect square and a perfect cube!  So we can use either pattern to factor! Both will work, but here's a tip: when you have a choice like this, use the difference of squares. It makes things easier if you do. So we will use difference of squares:
which. according to the pattern, factors into:
Will this factor further?  is obviously a perfect cube. Is 27 also a perfect cube? Answer: Yes, it is  . So we can use sum of cubes pattern on the first factor and the difference of cubes pattern on the second factor:
which simplifies to:
Nothing else will factor. So now we use the Zero Product Property:
 or  or  of
The first and third equations are easy to solve. The second and fourth ones require the Quadratic Formula. The solutions to the second and fourth equations work out to be complex numbers. We do not want complex number answers so we will reject those. The solutions to the frst and third equations are:
 or 
And since b is the base of a logarithm in the original equation and since bases of logarithms cannot be negative, we will reject b = -3 as a solution, too. So all that's left is:
b = 3
P.S. In theory, an equation like
has 6 solutions. By using the factoring method of solving we were able to find all six solutions: 2 real (3 and -3) and 4 complex. In this particular problem we ended up having reasons to reject 5 of the 6. But in other problems, you might need to use one or more of the 5 we rejected in this problem. By factoring we put ourselves in the position of having all possible solutions. If we had just found a 6th root we would have definitely missed the complex solutions and possibly the negative one, too. So it is safer to use factoring.
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