Answer
$x=0.600967$.
Work Step by Step
Given
$y=2^x$ and $y=1+3^{-x}$
From the graph, we can find the intersection point $(0.601,1.517)$
To find the intersection point using (Newton's Method), since
\begin{aligned}
f(x)& = 2^2-1-3^{-x}\\
f'(x)&= 2^x \ln 2+3^{-x} \ln 3
\end{aligned}
Then
\begin{aligned}
x_{n+1}&=x_n-\frac{f\left(x_n\right) }{ f^{\prime}\left(x_n\right)}\\
&= x_n- \frac{2^{x_n}-1-3^{-x_n}}{2^{x_n} \ln 2+3^{-x_n} \ln 3}
\end{aligned}
Using $x_0\approx 0.5$
Then
\begin{aligned}
x_1&= 0.5-\:\frac{2^{0.5}-1-3^{-0.5}}{2^{0.5}\:\ln \:2+3^{-0.5}\:\ln \:3}\approx 0.60104\\
x_2&\approx 0.600967\\
x_3&\approx 0.600967
\end{aligned}
So, correct to six decimal places, the root occurs at $x=0.600967$.
