SOLUTION: Let 2^a=5 and 2^b=7. Using exponent rules, solve the equations in terms of a and b. 0.4^x=49

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Let 2^a=5 and 2^b=7. Using exponent rules, solve the equations in terms of a and b. 0.4^x=49      Log On


   

Question 718404: Let 2^a=5 and 2^b=7. Using exponent rules, solve the equations in terms of a and b.
0.4^x=49
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
One way to solve this is...
  • Solve 2%5Ea=5 for a: a+=+log%282%2C+%285%29%29
  • Solve 2%5Eb=7 for b: b+=+log%282%2C+%287%29%29
  • Solve 0.4%5Ex=49 for x. Since and b are in terms of base 2 logarithms, we will use base 2 logs to solve for x:
    log%282%2C+%280.4%5Ex%29%29+=+log%282%2C+%2849%29%29
    x%2Alog%282%2C+%280.4%29%29+=+log%282%2C+%2849%29%29
    x+=+log%282%2C+%2849%29%29%2Flog%282%2C+%280.4%29%29
  • Now we try to rewrite log%282%2C+%2849%29%29%2Flog%282%2C+%280.4%29%29 in terms of a's and/or b's. Earlier we found that a+=+log%282%2C+%285%29%29 and b+=+log%282%2C+%287%29%29. So we can substitute a and b for these two logs. Plus, we should be able to figure out the base 2 log of any power of 2. So our task is to try to rewrite log%282%2C+%2849%29%29%2Flog%282%2C+%280.4%29%29 in terms of base 2 logs of 5's, 7's or any powers of 2. It's probably easy to see how to rewrite log%282%2C+%2849%29%29 (since 49 is 7 squared). The tricky part is log%282%2C+%280.4%29%29. If we rewrite 0.4 as a fraction the path will become clear: 0.4 = 4/10 = 2/5! We now have a path to a solution: