If the area above `x-`a xi s , bounded by the curves `y=2^(k x)a n dx=0a n dx=2i s3/((log)_e2),` then find the value of `k`

Explanation

Given that a region is bounded by the curves `y=2^(k x)`a n d`x=0`a n d`x=2`
Area of that region is given by,
`int_0^22^(k x)dx`
`=1/k[2^(kx)/(log 2)]_0^2`
`=[(2^(2k)-1)/(k log 2`
But `A=3/((log)_e2)`
`therefore [(2^(2k)-1)/(k log 2)]=3/((log)_e2)`
`=>2^(2k)-1=3k`
`=>4^k=3k+1`
this shows that `k=1`