Solve for x (e^4)^x*e^(x^2)=e^12

${(e^{4})}^{x} \cdot e^{x^{2}} = e^{12}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$e^{4 x} \cdot e^{x^{2}} = e^{12}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$e^{4 x + x^{2}} = e^{12}$

Since the bases are the same, then two expressions are only equal if the exponents are also equal.

$4 x + x^{2} = 12$

Solve for $x$ .

Subtract $12$ from both sides of the equation.

$4 x + x^{2} - 12 = 0$

Factor the left side of the equation.

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$4 u + u^{2} - 12 = 0$

Factor $4 u + u^{2} - 12$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 12$ and whose sum is $4$ .

$- 2, 6$

Write the factored form using these integers.

$(u - 2) (u + 6) = 0$

Replace all occurrences of $u$ with $x$ .

$(x - 2) (x + 6) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 2 = 0$

$x + 6 = 0$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Set $x + 6$ equal to $0$ and solve for $x$ .

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

The final solution is all the values that make $(x - 2) (x + 6) = 0$ true.

$x = 2, - 6$

Solve for x (e^4)^x*e^(x^2)=e^12

Related Questions

Support LCMGCF.com by disabling your adblocker.