SOLUTION: Let 2^a = 5 and 2^b = 9. Using exponent rules, solve the equation in terms of a and/or b. 5^x=32 I really don't even know what the question is asking me to do. Please help!

Algebra ->  Exponents -> SOLUTION: Let 2^a = 5 and 2^b = 9. Using exponent rules, solve the equation in terms of a and/or b. 5^x=32 I really don't even know what the question is asking me to do. Please help!       Log On


   

Question 275351: Let 2^a = 5 and 2^b = 9. Using exponent rules, solve the equation in terms of a and/or b.
5^x=32
I really don't even know what the question is asking me to do. Please help!
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
The goal here is to solve for 'x' in terms of either 'a' or 'b' (or both). In other words, have a solution of x = (something with 'a's and/or 'b's in it)


You have to play with it a bit, but it turns out that we never use 2%5Eb=9. So let's just work with 2%5Ea=5


First convert 2%5Ea=5 into logarithmic form to get a=log%282%2C%285%29%29. Now use the change of base formula to get a=log%2810%2C%285%29%29%2Flog%2810%2C%282%29%29


Now move onto 5%5Ex=32 and convert that into logarithmic form to get x=log%285%2C%2832%29%29. Rewrite 32 as 2%5E5 to get x=log%285%2C%282%5E5%29%29


Now pull down the exponent to get x=5%2Alog%285%2C%282%29%29 and use the change of base formula: x=5log%2810%2C%282%29%29%2Flog%2810%2C%285%29%29. Take note how this is very similar to a=log%2810%2C%285%29%29%2Flog%2810%2C%282%29%29 but this just has the reverse of that expression and an extra '5' in there. If we invert 'a', we then get 1%2Fa=log%2810%2C%282%29%29%2Flog%2810%2C%285%29%29 which is the missing piece.


So


So the solution, in terms of 'a', is x=5%2Fa