Question 204·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

In $2^{-x}$, a negative exponent means $1/2^x$, not a negative number. Changing $x$ to $-x$ swaps the terms in both $2^x+2^{-x}$ and $4^x+4^{-x}$. So a positive solution and its negative partner give the same requested sum.

Type $2^{x_1}+2^{-x_1}\sim10\{x_1>0\}$ as a regression, which asks Desmos to fit the unknown $x_1$. The $\sim$ tells Desmos to fit the two sides, and the restriction selects the positive solution. Under PARAMETERS, Desmos shows $x_1\approx3.307$.

Keep the regression and type $4^{x_1}+4^{-x_1}$ on the next line. Desmos uses the fitted value and prints $98$. The fitted $x_1$ isn't what you grid in; the value of $4^x+4^{-x}$ is $98$. Grid in 98.

Because $4=2^2$, the power-of-a-power rule turns each requested term into the square of a term in the given sum. Multiply the exponents:

The given sum equals $10$. Square both sides to bring those two squares into the equation:

Squaring a sum produces two middle products, not only the two squares. Multiply the two copies term by term:

When multiplying powers with the same base, add their exponents. A zero exponent makes a nonzero base equal $1$:

Replace the two squares with $4^x$ and $4^{-x}$, and replace their middle product with $1$:

Subtract the two middle products:

Type $10^2-2$ in Desmos. It prints $98$, so the requested sum is $98$. Grid in 98.