# Solve $5^{x^2+3x+2} \,=\, 1$

The given equation in this problem is an exponential equation basically but its exponent is a quadratic expression. So, it is a combination of the concepts of exponents and quadratics.

### Release the equation from exponentiation

In this exponential equation, the base of the exponential function is $5$ in the left-hand side of the equation but the right-hand side of the equation is a number and it is $1$. In order to solve this equation, the exponential equation should be released from the exponentiation. It is possible by expressing the right-hand side of the equation in exponential form as per zero exponent rule.

$\implies$ $5^{x^2+3x+2} \,=\, 5^0$

Now, compare the expressions of the both sides of the equation. In this step, the bases of the exponential expressions of both sides of the equation are same. So, the exponents should also be equal logically and it is acceptable by the mathematics.

$\implies$ $x^2+3x+2 \,=\, 0$

The given exponential equation is converted as a quadratic equation $x^2+3x+2 \,=\, 0$. Now, let us solve it for evaluating the variable $x$. The quadratic equation can be solved by factoring. It can be factored by splitting the middle term.

$\implies$ $x^2+2x+x+2 \,=\, 0$

$\implies$ $x(x+2)+1(x+2) \,=\, 0$

$\implies$ $(x+2)(x+1) \,=\, 0$

$\implies$ $x+1 \,=\, 0$ and $x+2 \,=\, 0$

Therefore $x \,=\, -1$ and $x \,=\, -2$ are the solution of the given exponential equation.

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