Solve the equation: 5^(x+1) = 26 − 5^(1-x)

We use the rules $a^{m+n} = a^m \times a^n$ and $a^{m-n} = a^m \times a^{-n} = \frac{a^m}{a^n}$.
So, $5^{x+1}$ can be written as $5^x \times 5^1 = 5 \times 5^x$.
And $5^{1-x}$ can be written as $5^1 \times 5^{-x} = \frac{5}{5^x}$.
Substituting these into the original equation:

$$5 \times 5^x = 26 - \frac{5}{5^x}$$

Let $y = 5^x$. Since $5^x$ is always positive, $y > 0$.
The equation becomes:

$$5y = 26 - \frac{5}{y}$$

Multiply the entire equation by $y$ to eliminate the fraction:

$$5y \times y = 26 \times y - \frac{5}{y} \times y$$ $$5y^2 = 26y - 5$$

Rearrange the terms to form a standard quadratic equation:

$$5y^2 - 26y + 5 = 0$$

We need two numbers that multiply to $5 \times 5 = 25$ and add up to $-26$. These numbers are $-1$ and $-25$.
Rewrite the middle term:

$$5y^2 - 25y - y + 5 = 0$$

Factor by grouping:

$$5y(y - 5) - 1(y - 5) = 0$$ $$(5y - 1)(y - 5) = 0$$

This gives two possible values for $y$:

$$5y - 1 = 0 \implies 5y = 1 \implies y = \frac{1}{5}$$ $$y - 5 = 0 \implies y = 5$$

Case 1: $y = \frac{1}{5}$

$$5^x = \frac{1}{5}$$

Since $\frac{1}{5} = 5^{-1}$, we have:

$$5^x = 5^{-1}$$

Therefore,

$$x = -1$$

Case 2: $y = 5$

$$5^x = 5$$

Since $5 = 5^1$, we have:

$$5^x = 5^1$$

Therefore,

$$x = 1$$

The solutions for $x$ are $-1$ and 1.

The final answers are $x=-1, x=1$.