Precalculus Examples

$e^{x^{2}} = e^{6 x} \cdot \frac{1}{e^{5}}$

Step 1

Move $e^{5}$ to the numerator using the negative exponent rule $\frac{1}{b^{n}} = b^{- n}$ .

$e^{x^{2}} = e^{6 x} \cdot e^{- 5}$

Step 2

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

$e^{x^{2}} = e^{6 x - 5}$

Step 3

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

$x^{2} = 6 x - 5$

Step 4

Solve for $x$ .

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Step 4.1

Subtract $6 x$ from both sides of the equation.

$x^{2} - 6 x = - 5$

Step 4.2

Add $5$ to both sides of the equation.

$x^{2} - 6 x + 5 = 0$

Step 4.3

Factor $x^{2} - 6 x + 5$ using the AC method.

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Step 4.3.1

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 $5$ and whose sum is $- 6$ .

$- 5, - 1$

Step 4.3.2

Write the factored form using these integers.

$(x - 5) (x - 1) = 0$

Step 4.4

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

$x - 5 = 0$

$x - 1 = 0$

Step 4.5

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

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Step 4.5.1

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

$x - 5 = 0$

Step 4.5.2

Add $5$ to both sides of the equation.

$x = 5$

Step 4.6

Set $x - 1$ equal to $0$ and solve for $x$ .

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Step 4.6.1

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 4.6.2

Add $1$ to both sides of the equation.

$x = 1$

Step 4.7

The final solution is all the values that make $(x - 5) (x - 1) = 0$ true.

$x = 5, 1$