Calculus Examples

$\frac{5^{x^{2}}}{125} = 25^{x}$

Step 1

Raise $125$ to the power of $1$ .

$\frac{5^{x^{2}}}{125^{1}} = 25^{x}$

Step 2

Move $125^{1}$ to the numerator using the negative exponent rule $\frac{1}{b^{n}} = b^{- n}$ .

$5^{x^{2}} \cdot 125^{- 1} = 25^{x}$

Step 3

Rewrite $125$ as $5^{3}$ .

$5^{x^{2}} \cdot {(5^{3})}^{- 1} = 25^{x}$

Step 4

Multiply the exponents in ${(5^{3})}^{- 1}$ .

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Step 4.1

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

$5^{x^{2}} \cdot 5^{3 \cdot - 1} = 25^{x}$

Step 4.2

Multiply $3$ by $- 1$ .

$5^{x^{2}} \cdot 5^{- 3} = 25^{x}$

Step 5

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

$5^{x^{2} - 3} = 25^{x}$

Step 6

Create equivalent expressions in the equation that all have equal bases.

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

Step 7

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

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

Step 8

Solve for $x$ .

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Step 8.1

Subtract $2 x$ from both sides of the equation.

$x^{2} - 3 - 2 x = 0$

Step 8.2

Factor $x^{2} - 3 - 2 x$ using the AC method.

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Step 8.2.1

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 $- 3$ and whose sum is $- 2$ .

$- 3, 1$

Step 8.2.2

Write the factored form using these integers.

$(x - 3) (x + 1) = 0$

Step 8.3

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

$x - 3 = 0$

$x + 1 = 0$

Step 8.4

Set $x - 3$ equal to $0$ and solve for $x$ .

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Step 8.4.1

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 8.4.2

Add $3$ to both sides of the equation.

$x = 3$

Step 8.5

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

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Step 8.5.1

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

$x + 1 = 0$

Step 8.5.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 8.6

The final solution is all the values that make $(x - 3) (x + 1) = 0$ true.

$x = 3, - 1$