Precalculus Examples

$\log (t + 3) + \log (t) = 1$

Step 1

Simplify the left side.

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Step 1.1

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

$\log ((t + 3) t) = 1$

Step 1.2

Apply the distributive property.

$\log (t \cdot t + 3 t) = 1$

Step 1.3

Multiply $t$ by $t$ .

$\log (t^{2} + 3 t) = 1$

Step 2

Rewrite $\log (t^{2} + 3 t) = 1$ 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 $b^{y} = x$ .

$10^{1} = t^{2} + 3 t$

Step 3

Solve for $t$ .

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Step 3.1

Rewrite the equation as $t^{2} + 3 t = 10^{1}$ .

$t^{2} + 3 t = 10$

Step 3.2

Subtract $10$ from both sides of the equation.

$t^{2} + 3 t - 10 = 0$

Step 3.3

Factor $t^{2} + 3 t - 10$ 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 $3$ .

$- 2, 5$

Step 3.3.2

Write the factored form using these integers.

$(t - 2) (t + 5) = 0$

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 $0$ .

$t - 2 = 0$

$t + 5 = 0$

Step 3.5

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

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Step 3.5.1

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

$t - 2 = 0$

Step 3.5.2

Add $2$ to both sides of the equation.

$t = 2$

Step 3.6

Set $t + 5$ equal to $0$ and solve for $t$ .

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Step 3.6.1

Set $t + 5$ equal to $0$ .

$t + 5 = 0$

Step 3.6.2

Subtract $5$ from both sides of the equation.

$t = - 5$

Step 3.7

The final solution is all the values that make $(t - 2) (t + 5) = 0$ true.

$t = 2, - 5$

Step 4

Exclude the solutions that do not make $\log (t + 3) + \log (t) = 1$ true.

$t = 2$