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}#

Describe your changes (optional) 200