# How do you solve 5^(x^2+2x)=125?

Dec 2, 2016

$x = - 3 , x = 1$

#### Explanation:

To solve the equation ${5}^{{x}^{2} + 2 x} = 125$ for variable $x$, we must apply the logarithm conversion formula.

logarithm to exponential conversion: $\setminus {\log}_{a} \left(b\right) = x \setminus \leftrightarrow {a}^{x} = b$

calculations

• ${5}^{{x}^{2} + 2 x} = 125 \setminus \Rightarrow \setminus {\log}_{5} \left(125\right) = {x}^{2} + 2 x$
• You're going to need a calculator to solve the left side's logarithm, but you should get the following:
$3 = {x}^{2} + 2 x$
• Now subtract 3 from both sides to form a quadratic equation:
$0 = {x}^{2} + 2 x - 3$
• You should get roots of -3 and 1. $\setminus \Rightarrow \left(x + 3\right) \left(x - 1\right)$

Therefore, $x = - 3 , 1$

Dec 4, 2016

$x = - 3$ or $x = 1$

#### Explanation:

Another way to approach this is to realize that $125 = {5}^{3}$. Then we see that

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

We now have two equal bases, each to a power. Since these are equal, we can say that their exponents must be equal. (We could write a rule for this and say that if ${a}^{b} = {a}^{c}$, then $b = c$).

So we know that

${x}^{2} + 2 x = 3$

${x}^{2} + 2 x - 3 = 0$
$\left(x + 3\right) \left(x - 1\right) = 0$
So $x = - 3$ or $x = 1$.