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

Question 975026: Help solve this equation
4^c=5(2^(c-1))+6
Answer by KMST(5347) (Show Source): You can put this solution on YOUR website!
is an exponential function,
with a shape like , always increasing.
So is , but does not increase as sharply,
because the exponent is half as much.
For , ,
but as increases, at some value of ,
eventually catches up with , and then surpasses it.
So there is one, and just one value of where ,
for lesser values of , ,
and for greater values of , .

Where is <---> ?
It is for some , but for what value of ?
How are you expected to find Out?
Guess and check? Graphic calculator?

<---><---><---><--->
If we only consider positive integer values of ,
We realize that is always an even number.
Also, as long as <--> , is even and is odd,
so we cannot get <---> with integers such that .
For to be even, must be odd,
and that only happens when <--><--><--><--> .
For that value of , , ,
and then <---> .
Since is the only integer solution of <---> ,
and because we knew that had one and just one solution,
is THE solution to .
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