Question 975026
{{{f(c)=4^c=(2^2)^c=2^2^(2c)}}} is an exponential function,
with a shape like {{{graph(300,300,-.2,.8,-.1,.9,0.1*2^(5x))}}} , always increasing.
So is {{{g(c)=5(2^(c-1))+6}}} , but {{{g(c)}}} does not increase as sharply,
because the exponent is half as much.
For {{{c=0}}} , {{{g(0)=5(2^(-1))+6=5(1/2)+6=5/2+6>h(0)=4^0=1}}} ,
but as {{{c}}} increases, at some value of {{{c>0}}} ,
{{{f(c)}}} eventually catches up with {{{g(c)}}} , and then surpasses it.
So there is one, and just one value of {{{c}}} where {{{f(c)=4^c=5(2^(c-1))+6=g(c)}}} ,
for lesser values of {{{c}}} , {{{f(c)<g(c)}}} ,
and for greater values of {{{c}}} , {{{f(c)>g(c)}}} .
 
Where is {{{f(c)=g(c)}}}<--->{{{4^c=5(2^(c-1))+6}}} ?
It is for some {{{c>0}}} , but for what value of {{{c>0}}} ?
How are you expected to find Out?
Guess and check? Graphic calculator?
 
{{{4^c=5(2^(c-1))+6}}}<--->{{{2^(2c)=5(2^(c-1))+6}}}<--->{{{2^(2c)/2=(5(2^(c-1))+6)/2}}}<--->{{{2^(2c-1)=5(2^(c-1))/2+6/2}}}<--->{{{2^(2c-1)=5(2^(c-2))+3}}}
If we only consider positive integer values of {{{c}}} ,
We realize that {{{2^(2c-1)}}} is always an even number.
Also, as long as {{{c-2>=1}}}<-->{{{c>=3}}} , {{{5(2^(c-2))}}} is even and {{{5(2^(c-2))+3}}} is odd,
so we cannot get {{{4^c=5(2^(c-1))+6}}}<--->{{{2^(2c-1)=5(2^(c-2))+3}}} with integers such that {{{c>=3}}}.
For {{{5(2^(c-2))+3}}} to be even, {{{5(2^(c-2))}}} must be odd,
and that only happens when {{{c=2}}}<-->{{{c-2=0}}}<-->{{{2^(c-2)=1}}}<-->{{{5(2^(c-2))=5}}}<-->{{{5(2^(c-2))+3=5+3=8}}} .
For that value of {{{c}}} , {{{c=2}}} , {{{2^(2c-1)=2^(2*2-1)=2^(4-1)=2^3=8}}} ,
and then {{{2^(2c-1)=5(2^(c-2))+3}}}<--->{{{4^c=5(2^(c-1))+6}}} .
Since {{{c=2}}} is the only integer solution of {{{4^c=5(2^(c-1))+6}}}<--->{{{2^(2c-1)=5(2^(c-2))+3}}} ,
and because we knew that {{{4^c=5(2^(c-1))+6}}} had one and just one solution,
{{{highlight(c=2)}}} is THE solution to {{{4^c=5(2^(c-1))+6}}} .