Solve for a log of a+ log of a+21=2

$\log (a) + \log (a + 21) = 2$

Step 1

Simplify the left side.

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

$\log (a (a + 21)) = 2$

Apply the distributive property.

$\log (a \cdot a + a \cdot 21) = 2$

Simplify the expression.

Multiply $a$ by $a$ .

$\log (a^{2} + a \cdot 21) = 2$

Move $21$ to the left of $a$ .

$\log (a^{2} + 21 a) = 2$

Step 2

Rewrite $\log (a^{2} + 21 a) = 2$ 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 $b^{y} = x$ .

$10^{2} = a^{2} + 21 a$

Step 3

Solve for $a$ .

Rewrite the equation as $a^{2} + 21 a = 10^{2}$ .

$a^{2} + 21 a = 10^{2}$

Raise $10$ to the power of $2$ .

$a^{2} + 21 a = 100$

Subtract $100$ from both sides of the equation.

$a^{2} + 21 a - 100 = 0$

Factor $a^{2} + 21 a - 100$ 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 $- 100$ and whose sum is $21$ .

$- 4, 25$

Write the factored form using these integers.

$(a - 4) (a + 25) = 0$

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

$a - 4 = 0$

$a + 25 = 0$

Set $a - 4$ equal to $0$ and solve for $a$ .

Set $a - 4$ equal to $0$ .

$a - 4 = 0$

Add $4$ to both sides of the equation.

$a = 4$

Set $a + 25$ equal to $0$ and solve for $a$ .

Set $a + 25$ equal to $0$ .

$a + 25 = 0$

Subtract $25$ from both sides of the equation.

$a = - 25$

The final solution is all the values that make $(a - 4) (a + 25) = 0$ true.

$a = 4, - 25$

Step 4

Exclude the solutions that do not make $\log (a) + \log (a + 21) = 2$ true.

$a = 4$

Solve for a log of a+ log of a+21=2

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