Solve 2^x + x = 5 for x.

Step-by-step solution

1. 1

   Recognise that no closed form exists. The unknown appears both in an exponent and as a linear term. Logarithms cannot separate them: taking $\ln$ of $2^{x}+x$ does not simplify, because the logarithm of a sum is not the sum of logarithms. So the honest route is numerical.

2. 2

   Confirm exactly one root and bracket it. Let $f(x)=2^{x}+x-5$. Both $2^{x}$ and $x$ are strictly increasing, so $f$ is strictly increasing and can cross zero at most once. Evaluating:

   $f(1)=2+1-5=-2<0,\qquad f(2)=4+2-5=1>0$

   A sign change on a continuous function guarantees a root in $(1,2)$, and monotonicity guarantees it is the only one.

3. 3

   Narrow the bracket by testing.

   $f(1.5)=2^{1.5}+1.5-5\approx2.8284+1.5-5=-0.6716<0$

   $f(1.7)\approx3.2490+1.7-5=-0.0510<0,\qquad f(1.72)\approx3.2943+1.72-5=+0.0143>0$

   so the root lies between $1.70$ and $1.72$, already close to $1.72$.

4. 4

   Bisect to convergence. Repeatedly halving the bracket $[1.7,1.72]$ and keeping the half where the sign changes converges to

   $x\approx1.7156207$

   Check: $2^{1.7156207}+1.7156207=5.0000000$ to within $10^{-12}$ ✓.

5. 5

   Round carefully — the third decimal matters. To three decimals the root is $x\approx1.716$, not $1.717$: substituting $1.717$ gives $2^{1.717}+1.717\approx5.0041$, which overshoots $5$ by about $0.004$. Because $f$ climbs at roughly $2.9$ per unit near the root, an error of $0.001$ in $x$ shifts $f$ by about $0.003$ — enough to see, so the extra digit is worth carrying.