SOLUTION: If 2^a = 3, 3^b = 2, find 1/(a+1) + 1/(b+1)

Algebra ->  Test -> SOLUTION: If 2^a = 3, 3^b = 2, find 1/(a+1) + 1/(b+1)       Log On


   

Question 1209572: If 2^a = 3, 3^b = 2,
find 1/(a+1) + 1/(b+1)
Found 3 solutions by ikleyn, greenestamps, mccravyedwin:
Answer by ikleyn(52778) About Me  (Show Source):
You can put this solution on YOUR website!
.
If 2^a = 3, 3^b = 2,
find 1/(a+1) + 1/(b+1)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

If 2^a = 3  and  3^b = 2,  then  2^(ab) = (2^a)^b = 3^b = 2,  which implies  ab = 1.


Now,    1%2F%28a%2B1%29 + 1%2F%28b%2B1%29 = %28%28a%2B1%29+%2B+%28b%2B1%29%29%2F%28%28a%2B1%29%2A%28b%2B1%29%29 = 

      = %28a+%2B+1+%2B+b+%2B+1%29%2F%28ab+%2B+a+%2B+b+%2B+1%29 = %28a%2Bb%2B2%29%2F%281+%2B+a+%2B+b+%2B+1%29 = %28a%2Bb%2B2%29%2F%28a%2Bb%2B2%29 = 1.


ANSWER.  If  2^a = 3,  3^b = 2,  then  1%2F%28a%2B1%29 + 1%2F%28b%2B1%29 = 1.

At this point, the problem is solved to the end.

The answer is really unexpected.

True / good entertainment problem with the unexpected end.



Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


Here is a very different solution. More work than the solution provided by tutor @ikleyn... but it gives the student a lot of practice with rules of logarithms.

2%5Ea=3 --> a=log%282%2C3%29

3%5Eb=2 --> b=log%283%2C2%29





1%2F%28a%2B1%29%2B1%2F%28b%2B1%29=log%286%2C2%29%2Blog%286%2C3%29=log%286%2C6%29=1

ANSWER: 1

Answer by mccravyedwin(406) About Me  (Show Source):
You can put this solution on YOUR website!

That's a clever way for it avoids logarithms. But I think this student is
studying solving exponential equations by taking natural logs of both sides.  
So I think the student would do it this way:

                 

Edwin