Solve for x e^(x^2)=e^(11x)*1/(e^30)

$e^{x^{2}} = e^{11 x} \cdot \frac{1}{e^{30}}$

Step 1

Move $e^{30}$ to the numerator using the negative exponent rule $\frac{1}{b^{n}} = b^{- n}$ .

$e^{x^{2}} = e^{11 x} \cdot e^{- 30}$

Step 2

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

$e^{x^{2}} = e^{11 x - 30}$

Step 3

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

$x^{2} = 11 x - 30$

Step 4

Solve for $x$ .

Subtract $11 x$ from both sides of the equation.

$x^{2} - 11 x = - 30$

Add $30$ to both sides of the equation.

$x^{2} - 11 x + 30 = 0$

Factor $x^{2} - 11 x + 30$ 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 $30$ and whose sum is $- 11$ .

$- 6, - 5$

Write the factored form using these integers.

$(x - 6) (x - 5) = 0$

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

$x - 6 = 0$

$x - 5 = 0$

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

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

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

The final solution is all the values that make $(x - 6) (x - 5) = 0$ true.

$x = 6, 5$

Solve for x e^(x^2)=e^(11x)*1/(e^30)

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