What are the two solutions for the equation: x+2=2^x?

Question

Pre-Calc using logs

The textbook claims there are 2 answers, but I only can find 1...2 because 2 + 2 = 4 and 2^2 =4.  I have tried logs as well as guess and check with whole and rational numbers.  What is the second solution?

Answer 1

Yes, there are two solutions.

x=2 is, as you said, one solution. Newton's Method can be used to find the other one.

2^x=e^xln(2), (d/dx)2^x=2^xe^xln(2).

Let f(x)=x+2-2^x, f'(x)=1-2^xln(2).

x_n+1=x_n-f(x_n)/f'(x_n). Let x₀=0.

x₁=-1/(1-ln(2))=-3.258...

x₂=-1.789...

x₃=-1.690...

x₄=-1.69009...

x₅=-1.690093068...

x₆=-1.69009306762...

So the two solutions are 2 (as expected) and -1.6901 approx.

Newton's Method can be used to find x=2 if we start with x₀=1.

The graph below shows why there are two solutions (x-intercepts):

GRAPH OF x+2-2^x

Courtesy of Desmos.com.