Solution

Let

$f(2^k) = g(k)$

Then recurrence becomes

$g(k) = 5g(k-1) + 3$

with

$g(0) = f(1) = 7$

Solve recurrence.

Homogeneous solution:

$g_h(k) = C5^k$

Particular solution:

Assume constant $A$

$A = 5A + 3$

$-4A = 3$

$A = -\frac{3}{4}$

General solution:

$g(k) = C5^k - \frac{3}{4}$

Use initial condition

$g(0) = 7$

$C - \frac{3}{4} = 7$

$C = \frac{31}{4}$

Thus

$f(2^k) = \frac{31}{4}5^k - \frac{3}{4}$