SOLUTION: Help solve this equation 4^c=5(2^(c-1))+6

Algebra ->  Rational-functions -> SOLUTION: Help solve this equation 4^c=5(2^(c-1))+6      Log On


   

Question 975026: Help solve this equation
4^c=5(2^(c-1))+6
Answer by KMST(5347) About Me  (Show Source):
You can put this solution on YOUR website!
f%28c%29=4%5Ec=%282%5E2%29%5Ec=2%5E2%5E%282c%29 is an exponential function,
with a shape like graph%28300%2C300%2C-.2%2C.8%2C-.1%2C.9%2C0.1%2A2%5E%285x%29%29 , always increasing.
So is g%28c%29=5%282%5E%28c-1%29%29%2B6 , but g%28c%29 does not increase as sharply,
because the exponent is half as much.
For c=0 , g%280%29=5%282%5E%28-1%29%29%2B6=5%281%2F2%29%2B6=5%2F2%2B6%3Eh%280%29=4%5E0=1 ,
but as c increases, at some value of c%3E0 ,
f%28c%29 eventually catches up with g%28c%29 , and then surpasses it.
So there is one, and just one value of c where f%28c%29=4%5Ec=5%282%5E%28c-1%29%29%2B6=g%28c%29 ,
for lesser values of c , f%28c%29%3Cg%28c%29 ,
and for greater values of c , f%28c%29%3Eg%28c%29 .

Where is f%28c%29=g%28c%29<--->4%5Ec=5%282%5E%28c-1%29%29%2B6 ?
It is for some c%3E0 , but for what value of c%3E0 ?
How are you expected to find Out?
Guess and check? Graphic calculator?

4%5Ec=5%282%5E%28c-1%29%29%2B6<--->2%5E%282c%29=5%282%5E%28c-1%29%29%2B6<--->2%5E%282c%29%2F2=%285%282%5E%28c-1%29%29%2B6%29%2F2<--->2%5E%282c-1%29=5%282%5E%28c-1%29%29%2F2%2B6%2F2<--->2%5E%282c-1%29=5%282%5E%28c-2%29%29%2B3
If we only consider positive integer values of c ,
We realize that 2%5E%282c-1%29 is always an even number.
Also, as long as c-2%3E=1<-->c%3E=3 , 5%282%5E%28c-2%29%29 is even and 5%282%5E%28c-2%29%29%2B3 is odd,
so we cannot get 4%5Ec=5%282%5E%28c-1%29%29%2B6<--->2%5E%282c-1%29=5%282%5E%28c-2%29%29%2B3 with integers such that c%3E=3.
For 5%282%5E%28c-2%29%29%2B3 to be even, 5%282%5E%28c-2%29%29 must be odd,
and that only happens when c=2<-->c-2=0<-->2%5E%28c-2%29=1<-->5%282%5E%28c-2%29%29=5<-->5%282%5E%28c-2%29%29%2B3=5%2B3=8 .
For that value of c , c=2 , 2%5E%282c-1%29=2%5E%282%2A2-1%29=2%5E%284-1%29=2%5E3=8 ,
and then 2%5E%282c-1%29=5%282%5E%28c-2%29%29%2B3<--->4%5Ec=5%282%5E%28c-1%29%29%2B6 .
Since c=2 is the only integer solution of 4%5Ec=5%282%5E%28c-1%29%29%2B6<--->2%5E%282c-1%29=5%282%5E%28c-2%29%29%2B3 ,
and because we knew that 4%5Ec=5%282%5E%28c-1%29%29%2B6 had one and just one solution,
highlight%28c=2%29 is THE solution to 4%5Ec=5%282%5E%28c-1%29%29%2B6 .