Basic Math Examples

$\log (y) + \log (y - 9) = 1$

Step 1

Simplify the left side.

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Step 1.1

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\log (y (y - 9)) = 1$

Step 1.2

Apply the distributive property.

$\log (y \cdot y + y \cdot - 9) = 1$

Step 1.3

Simplify the expression.

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Step 1.3.1

Multiply $y$ by $y$ .

$\log (y^{2} + y \cdot - 9) = 1$

Step 1.3.2

Move $- 9$ to the left of $y$ .

$\log (y^{2} - 9 y) = 1$

Step 2

Rewrite $\log (y^{2} - 9 y) = 1$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$10^{1} = y^{2} - 9 y$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $y^{2} - 9 y = 10^{1}$ .

$y^{2} - 9 y = 10$

Step 3.2

Subtract $10$ from both sides of the equation.

$y^{2} - 9 y - 10 = 0$

Step 3.3

Factor $y^{2} - 9 y - 10$ using the AC method.

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Step 3.3.1

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 $- 10$ and whose sum is $- 9$ .

$- 10, 1$

Step 3.3.2

Write the factored form using these integers.

$(y - 10) (y + 1) = 0$

Step 3.4

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

$y - 10 = 0$

$y + 1 = 0$

Step 3.5

Set $y - 10$ equal to $0$ and solve for $y$ .

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Step 3.5.1

Set $y - 10$ equal to $0$ .

$y - 10 = 0$

Step 3.5.2

Add $10$ to both sides of the equation.

$y = 10$

Step 3.6

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

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Step 3.6.1

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

$y + 1 = 0$

Step 3.6.2

Subtract $1$ from both sides of the equation.

$y = - 1$

Step 3.7

The final solution is all the values that make $(y - 10) (y + 1) = 0$ true.

$y = 10, - 1$

Step 4

Exclude the solutions that do not make $\log (y) + \log (y - 9) = 1$ true.

$y = 10$