SOLUTION: How would you do a problem like, (3^x)/((3^x)+3)=1/3

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: How would you do a problem like, (3^x)/((3^x)+3)=1/3      Log On


   

Question 251139: How would you do a problem like, (3^x)/((3^x)+3)=1/3
Found 2 solutions by jsmallt9, solver91311:
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
It might be easier to understand the solution if we break it into two parts:
  1. Figure out what 3%5Ex is
  2. Use part 1 to find what x is

To solve for 3%5Ex I am going to use a temporary variable:
Let q+=+3%5Ex
Now our equation is:
q%2F%28q%2B3%29+=+1%2F3
We can eliminate the fractions by multiplying by the lowest common denominator (which is 3*(q+3)):
3%2A%28q%2B3%29%28q%2F%28q%2B3%29%29+=+3%2A%28q%2B3%29%281%2F3%29
On the left the (q+3)'s cancel and on the right the 3's cancel leaving:
3%2Aq+=+%28q%2B3%29%2A1
or
3q+=+q%2B+3
Subtract q from each side:
2q+=+3
Divide both sides by 2:
q+=+3%2F2
Replace q with 3%5Ex
3%5Ex+=+3%2F2
(We could have solved for 3%5Ex without the temporary variable but most students find it a little harder to follow.) Now that we know what 3%5Ex is we can solve for x. Since 3/2 is not an obvious power of 3 and since x is in an exponent we will use logarithms. If you do not need any decimal approximation for the answer then base 3 logarithms would be best. (I'll show you later.) In case you do need a decimal approximation use a logarithm your calculator can find, like base 10:
log%28%283%5Ex%29%29+=+log%28%283%2F2%29%29
Now we can use a property of logarithms, log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29, to move the exponent of the argument out in front. This is one way to get variables out of exponents and the reason we use logarithms to solve these equations:
x%2Alog%28%283%29%29+=+log%28%283%2F2%29%29
Divide both sides by log(3):
x+=+log%28%283%2F2%29%29%2Flog%28%283%29%29
This is an exact answer for x. If you need a decimal approximation for the answer then find the two logarithms and then divide them.

If you use base 3 logarithms we would end up with:
x+=+log%283%2C+%283%2F2%29%29%2Flog%283%2C+%283%29%29
Since log%283%2C+%283%29%29+=+1 by definition we get:
x+=+log%283%2C+%283%2F2%29%29%2F1
which is
x+=+log%283%2C+%283%2F2%29%29
And this will simplify even further if we use another property of logarithms, log%28a%2C+%28p%2Fq%29%29+=+log%28a%2C+%28p%29%29+-+log%28a%2C+%28q%29%29:
x+=+log%283%2C+%283%29%29+-+log%283%2C+%282%29%29
x+=+1+-+log%283%2C+%282%29%29
which is a much simpler exact answer. But getting a decimal approximation for this would be a bit of a pain since there are no calculators I know of which can find base 3 logarithms. (You could use the base conversion formula, however.)

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!




Factor a 3 out of the LHS denominator:



Note the factor of 3 in both denominators; eliminate it:



Multiply by the remaining denominator:



Add to both sides:



Factor out of the LHS:









Take the log of both sides. Use base 3 logs:



Use





Then use





Is the exact answer. If you need a numerical approximation, use base conversion:



and do a little calculator work.

Late update: JSmall's idea of simplifying a step further by using the difference of the logs is an outstanding one. Use it. Also, if the first part of my explanation is too hard to follow, look at his substitution method -- much easier.

John