Precalculus Examples

$\log_{5} (3 x) + \log_{5} (2 x - 1) = \log_{5} (16 x - 10)$

Step 1

Simplify the left side.

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Step 1.1

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

$\log_{5} (3 x (2 x - 1)) = \log_{5} (16 x - 10)$

Step 1.2

Simplify by multiplying through.

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Step 1.2.1

Apply the distributive property.

$\log_{5} (3 x (2 x) + 3 x \cdot - 1) = \log_{5} (16 x - 10)$

Step 1.2.2

Simplify the expression.

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Step 1.2.2.1

Rewrite using the commutative property of multiplication.

$\log_{5} (3 \cdot (2 x \cdot x) + 3 x \cdot - 1) = \log_{5} (16 x - 10)$

Step 1.2.2.2

Multiply $- 1$ by $3$ .

$\log_{5} (3 \cdot (2 x \cdot x) - 3 x) = \log_{5} (16 x - 10)$

Step 1.3

Simplify each term.

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Step 1.3.1

Multiply $x$ by $x$ by adding the exponents.

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Step 1.3.1.1

Move $x$ .

$\log_{5} (3 \cdot (2 (x \cdot x)) - 3 x) = \log_{5} (16 x - 10)$

Step 1.3.1.2

Multiply $x$ by $x$ .

$\log_{5} (3 \cdot (2 x^{2}) - 3 x) = \log_{5} (16 x - 10)$

Step 1.3.2

Multiply $3$ by $2$ .

$\log_{5} (6 x^{2} - 3 x) = \log_{5} (16 x - 10)$

Step 2

For the equation to be equal, the argument of the logarithms on both sides of the equation must be equal.

$6 x^{2} - 3 x = 16 x - 10$

Step 3

Solve for $x$ .

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Step 3.1

Move all terms containing $x$ to the left side of the equation.

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Step 3.1.1

Subtract $16 x$ from both sides of the equation.

$6 x^{2} - 3 x - 16 x = - 10$

Step 3.1.2

Subtract $16 x$ from $- 3 x$ .

$6 x^{2} - 19 x = - 10$

Step 3.2

Add $10$ to both sides of the equation.

$6 x^{2} - 19 x + 10 = 0$

Step 3.3

Factor by grouping.

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Step 3.3.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 6 \cdot 10 = 60$ and whose sum is $b = - 19$ .

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Step 3.3.1.1

Factor $- 19$ out of $- 19 x$ .

$6 x^{2} - 19 x + 10 = 0$

Step 3.3.1.2

Rewrite $- 19$ as $- 4$ plus $- 15$

$6 x^{2} + (- 4 - 15) x + 10 = 0$

Step 3.3.1.3

Apply the distributive property.

$6 x^{2} - 4 x - 15 x + 10 = 0$

Step 3.3.2

Factor out the greatest common factor from each group.

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Step 3.3.2.1

Group the first two terms and the last two terms.

$(6 x^{2} - 4 x) - 15 x + 10 = 0$

Step 3.3.2.2

Factor out the greatest common factor (GCF) from each group.

$2 x (3 x - 2) - 5 (3 x - 2) = 0$

Step 3.3.3

Factor the polynomial by factoring out the greatest common factor, $3 x - 2$ .

$(3 x - 2) (2 x - 5) = 0$

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 $0$ .

$3 x - 2 = 0$

$2 x - 5 = 0$

Step 3.5

Set $3 x - 2$ equal to $0$ and solve for $x$ .

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Step 3.5.1

Set $3 x - 2$ equal to $0$ .

$3 x - 2 = 0$

Step 3.5.2

Solve $3 x - 2 = 0$ for $x$ .

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Step 3.5.2.1

Add $2$ to both sides of the equation.

$3 x = 2$

Step 3.5.2.2

Divide each term in $3 x = 2$ by $3$ and simplify.

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Step 3.5.2.2.1

Divide each term in $3 x = 2$ by $3$ .

$\frac{3 x}{3} = \frac{2}{3}$

Step 3.5.2.2.2

Simplify the left side.

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Step 3.5.2.2.2.1

Cancel the common factor of $3$ .

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Step 3.5.2.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{2}{3}$

Step 3.5.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{2}{3}$

Step 3.6

Set $2 x - 5$ equal to $0$ and solve for $x$ .

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Step 3.6.1

Set $2 x - 5$ equal to $0$ .

$2 x - 5 = 0$

Step 3.6.2

Solve $2 x - 5 = 0$ for $x$ .

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Step 3.6.2.1

Add $5$ to both sides of the equation.

$2 x = 5$

Step 3.6.2.2

Divide each term in $2 x = 5$ by $2$ and simplify.

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Step 3.6.2.2.1

Divide each term in $2 x = 5$ by $2$ .

$\frac{2 x}{2} = \frac{5}{2}$

Step 3.6.2.2.2

Simplify the left side.

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Step 3.6.2.2.2.1

Cancel the common factor of $2$ .

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Step 3.6.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{5}{2}$

Step 3.6.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{5}{2}$

Step 3.7

The final solution is all the values that make $(3 x - 2) (2 x - 5) = 0$ true.

$x = \frac{2}{3}, \frac{5}{2}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \frac{2}{3}, \frac{5}{2}$

Decimal Form:

$x = 0.\overline{6}, 2.5$

Mixed Number Form:

$x = \frac{2}{3}, 2 \frac{1}{2}$