Calculus Examples

$x^{2} - x = \log_{5} (25)$

Step 1

Logarithm base $5$ of $25$ is $2$ .

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Step 1.1

Rewrite as an equation.

$x^{2} - x = \log_{5} (25) = x$

Step 1.2

Rewrite $\log_{5} (25) = x$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b$ does not equal $1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

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

Step 1.3

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

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

Step 1.4

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

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

Step 1.5

The variable $x$ is equal to $2$ .

$x^{2} - x = 2$

Step 2

Subtract $2$ from both sides of the equation.

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

Step 3

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

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Step 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 $- 2$ and whose sum is $- 1$ .

$- 2, 1$

Step 3.2

Write the factored form using these integers.

$(x - 2) (x + 1) = 0$

Step 4

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

$x - 2 = 0$

$x + 1 = 0$

Step 5

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

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Step 5.1

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

$x - 2 = 0$

Step 5.2

Add $2$ to both sides of the equation.

$x = 2$

Step 6

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

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Step 6.1

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

$x + 1 = 0$

Step 6.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 7

The final solution is all the values that make $(x - 2) (x + 1) = 0$ true.

$x = 2, - 1$