Solve for x log of 6x+5- log of 3 = log of 2- log of x

$\log (6 x + 5) - \log (3) = \log (2) - \log (x)$

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\log (\frac{6 x + 5}{3}) = \log (2) - \log (x)$

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\log (\frac{6 x + 5}{3}) = \log (\frac{2}{x})$

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

$\frac{6 x + 5}{3} = \frac{2}{x}$

Solve for $x$ .

Multiply the numerator of the first fraction by the denominator of the second fraction. Set this equal to the product of the denominator of the first fraction and the numerator of the second fraction.

$(6 x + 5) x = 3 \cdot 2$

Solve the equation for $x$ .

Simplify $(6 x + 5) x$ .

Apply the distributive property.

$6 x \cdot x + 5 x = 3 \cdot 2$

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

Move $x$ .

$6 (x \cdot x) + 5 x = 3 \cdot 2$

Multiply $x$ by $x$ .

$6 x^{2} + 5 x = 3 \cdot 2$

Multiply $3$ by $2$ .

$6 x^{2} + 5 x = 6$

Subtract $6$ from both sides of the equation.

$6 x^{2} + 5 x - 6 = 0$

Factor by grouping.

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 $a \cdot c = 6 \cdot - 6 = - 36$ and whose sum is $b = 5$ .

Factor $5$ out of $5 x$ .

$6 x^{2} + 5 (x) - 6 = 0$

Rewrite $5$ as $- 4$ plus $9$

$6 x^{2} + (- 4 + 9) x - 6 = 0$

Apply the distributive property.

$6 x^{2} - 4 x + 9 x - 6 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(6 x^{2} - 4 x) + 9 x - 6 = 0$

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

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

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

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

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

$3 x - 2 = 0$

$2 x + 3 = 0$

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

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

$3 x - 2 = 0$

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

Add $2$ to both sides of the equation.

$3 x = 2$

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

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

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

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

$2 x + 3 = 0$

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

Subtract $3$ from both sides of the equation.

$2 x = - 3$

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

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

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

Exclude the solutions that do not make $\log (6 x + 5) - \log (3) = \log (2) - \log (x)$ true.

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

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = 0.\overline{6}$

Solve for x log of 6x+5- log of 3 = log of 2- log of x

Related Questions

Support LCMGCF.com by disabling your adblocker.