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Solve this equation

$4^{1+x} + 4^{1-x} = 10$

2. $\displaystyle 4^{1+x} + 4^{1-x} = 10$

$\displaystyle 4\cdot 4^{x} + 4\cdot 4^{-x} = 10$

$\displaystyle 4^x + 4^{-x} = \frac{5}{2}$

$\displaystyle 4^x(4^x + 4^{-x}) = \frac{5}{2}(4^x)$

$\displaystyle (4^{x})^2 + 1 = \frac{5}{2}(4^x)$

$\displaystyle (4^x)^2 - \frac{5}{2}(4^x) + 1 = 0$.

Now let $\displaystyle X = 4^x$ to get

$\displaystyle X^2 - \frac{5}{2}X + 1 = 0$

$\displaystyle 2X^2 - 5X + 2 = 0$

$\displaystyle 2X^2 - X - 4X + 2 = 0$

$\displaystyle X(2X-1) - 2(2X-1) =0$

$\displaystyle (2X-1)(X-2)=0$

$\displaystyle 2X-1 = 0$ or $\displaystyle X-2=0$

$\displaystyle X=\frac{1}{2}$ or $\displaystyle X=2$

$\displaystyle 4^x = \frac{1}{2}$ or $\displaystyle 4^x = 2$

$\displaystyle 2^{2x} = 2^{-1}$ or $\displaystyle 2^{2x} = 2^1$

$\displaystyle 2x = -1$ or $\displaystyle 2x = 1$

$\displaystyle x = -\frac{1}{2}$ or $\displaystyle x=\frac{1}{2}$.

3. Using hyperbolic cosine:

$4^{1+x} + 4^{1-x} = 10$

$4(4^{x} + 4^{-x}) = 10$

$4^{x} + 4^{-x} = \frac{5}{2}$

$(e^{\ln(4)})^x + (e^{\ln(4)})^{-x} = \frac{5}{2}$

$e^{x \ln(4)} + e^{-x \ln(4)} = \frac{5}{2}$

$\displaystyle \frac{e^{x \ln(4)} + e^{-x \ln(4)}}{2} = \frac{5}{4}$

$\cosh(x \ln(4)) = \frac{5}{4}$

$x \ln(4) = \pm \cosh^{-1}(\frac{5}{4})$

$x = \pm \cosh^{-1}(\frac{5}{4}) / \ln(4)$

$x = \pm \frac{1}{2}$