Exponential equation reducible to quadratic form

Examples

Here's an example of solving a quadratic equation using the quadratic formula. Solve the equation: $5^{x+1}+5^{2-x}=5^3+1$

Explanation:

Given equation: $5^{x+1}+5^{2-x}=5^3+1$

Step 1: Simplify the left-hand side by combining like terms:

$5^{x+1}+5^{2-x}=125+1$

$5^{x+1}+5^{2-x}=126$

Step 2: Write the powers of $5$ in a common base, which is $5$, and simplify:

$5 \cdot 5^x + 5^{-x} \cdot 5^2 = 126$

$5^{x+1} + 5^{2-x} = 126$

Step 3: Let $y=5^x$ to get a quadratic equation in terms of $y$:

$5y + \frac{25}{y} = 126$

Multiplying both sides by $y$ gives:

$5y^2 + 25 = 126y$

$5y^2 - 126y + 25 = 0$

Step 4: Factor the quadratic equation or quadratic formula can be applied:

$5y^2 - 126y + 25 = (5y-1)(y-25) = 0$

Step 5: Solve for $y$ by setting each factor equal to zero:

$5y-1=0 \space or \space y-25=0$

$y=\frac{1}{5}$ or $y=25$

Step 6: Substitute back $5^x$ for $y$ and solve for $x$:

$5^x=\frac{1}{5}$ or $5^x=25$

$5^x=5^{-1}$ or $5^x=5^2$

$x=-1$ or $x=2$

Step 7: Check the solutions:

Substituting $x=-1$ into the original equation, we get:

$5^{(-1)+1} + 5^{2-(-1)} = 5^3+1$

$5^0 + 5^3 = 125+1$

$1 + 125 = 126$

The left-hand side equals the right-hand side, so $x=-1$ is a valid solution.

Substituting $x=2$ into the original equation, we get:

$5^{(2)+1} + 5^{2-(2)} = 5^3+1$

$5^3 + 5^0 = 125+1$

$125 + 1 = 126$

The left-hand side equals the right-hand side, so $x=2$ is also a valid solution.

Solution:

$5^{x+1}+5^{2-x}=5^3+1$

$5^{x+1}+5^{2-x}=125+1$ $5^{x+1}+5^{2-x}=126$ $5^1.5^x+5^2.5^{-x}=126$ $5^1.5^x+\frac{5^2}{5^{x}}=126$

Let $5^x=y$ $5y+\frac{25}{y}=126$ $5y^2+\frac{25y}{y}=126y$ $5y^2-126y+25=0$

Here, $a = 5, b = -126,$ and $c = 25$ Quadratic Formula: $x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Substitute the values of $a, b,$ and $c$ in the quadratic formula:

$y=\frac{-(-126)\pm\sqrt{(-126)^2-4(5)(25)}}{2(5)}$

Simplifying: $y=\frac{126\pm\sqrt{15876-500}}{10}$

$y=\frac{126\pm\sqrt{15376}}{10}$

$y=\frac{126\pm124}{10}$ $y=\frac{250}{10}$; $y=\frac{2}{10}\\$$y=25$ ; $y=\frac{1}{5}$

Substituting back $y=5^x$:

$5^x=25$ or $5^x=\frac15$

$5^x=5^2$ or $5^x=5^{-1}$

$x=2$ or $x=-1$

Therefore, the solutions to the equation $5^{x+1}+5^{2-x}=5^3+1$ are

$x = 2$ and $x= -1$

The solution set of the quadratic equation is $\{2,-1\}$