We will develop an algebraic method instead.
#2^(x+6)=2x+15#
This is an equation of type
#a b^x+c x+d=0#
that can be solved using the so called Lambert function.
https://en.wikipedia.org/wiki/Lambert_W_function
We will make transformations in order to obtain an equation of type
#(c_1 y)e^(c_1 y) = c_2# obtaining the solution as
#c_1 y = W(c_2)# and then
#y = (W(c_2))/c_1#
I) Obtaining #(c_1 y)e^(c_1 y) = c_2# from #a b^x+c x+d=0#
making #b^x = e^(x log b)# and #y = -(cx+d)# we get
#a e^(-((y-d)/c) log b) = y# or
#a e^((d/c)logb) e^(-(logb/c)y) = y# or
#a b^(d/c)e^(-(logb/c)y) = y# or
#a b^(d/c) = y e^((logb/c)y) # or
#a b^(d/c) logb/c = (log b/c y) e^((logb/c)y)#
then here
#c_1 = logb/c# and #c_2 = a b^(d/c) logb/c = a b^(d/c) c_1#
and then
#y = -(cx+d) = (W(a b^(d/c) logb/c ))/(( logb/c))# and finally
#x = -1/c( (W(a b^(d/c) logb/c ))/(( logb/c))+d)#
Now substituting
#a = 2^6, b = 2, c = -2,d = -15#
we obtain
#x =- (15 log2 +2 W(-log2/(4 sqrt[2])))/(2 log2)# with two solutions
#x = {-7.2964344147839,-2.75297694275627}#
