Solve for x log base 16 of 9x+5- log base 16 of x^2-1=1/2

$\log_{16} (9 x + 5) - \log_{16} (x^{2} - 1) = \frac{1}{2}$

Simplify the left side.

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

$\log_{16} (\frac{9 x + 5}{x^{2} - 1}) = \frac{1}{2}$

Simplify the denominator.

Rewrite $1$ as $1^{2}$ .

$\log_{16} (\frac{9 x + 5}{x^{2} - 1^{2}}) = \frac{1}{2}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$\log_{16} (\frac{9 x + 5}{(x + 1) (x - 1)}) = \frac{1}{2}$

Rewrite $\log_{16} (\frac{9 x + 5}{(x + 1) (x - 1)}) = \frac{1}{2}$ 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 $b^{y} = x$ .

$16^{\frac{1}{2}} = \frac{9 x + 5}{(x + 1) (x - 1)}$

Cross multiply to remove the fraction.

$9 x + 5 = 16^{\frac{1}{2}} ((x + 1) (x - 1))$

Simplify $16^{\frac{1}{2}} ((x + 1) (x - 1))$ .

Reduce the expression by cancelling the common factors.

Rewrite $16$ as $4^{2}$ .

$9 x + 5 = {(4^{2})}^{\frac{1}{2}} ((x + 1) (x - 1))$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$9 x + 5 = 4^{2 (\frac{1}{2})} ((x + 1) (x - 1))$

Cancel the common factor of $2$ .

Cancel the common factor.

$9 x + 5 = 4^{2 (\frac{1}{2})} ((x + 1) (x - 1))$

Rewrite the expression.

$9 x + 5 = 4^{1} ((x + 1) (x - 1))$

Evaluate the exponent.

$9 x + 5 = 4 ((x + 1) (x - 1))$

Expand $(x + 1) (x - 1)$ using the FOIL Method.

Apply the distributive property.

$9 x + 5 = 4 (x (x - 1) + 1 (x - 1))$

Apply the distributive property.

$9 x + 5 = 4 (x \cdot x + x \cdot - 1 + 1 (x - 1))$

Apply the distributive property.

$9 x + 5 = 4 (x \cdot x + x \cdot - 1 + 1 x + 1 \cdot - 1)$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$9 x + 5 = 4 (x^{2} + x \cdot - 1 + 1 x + 1 \cdot - 1)$

Move $- 1$ to the left of $x$ .

$9 x + 5 = 4 (x^{2} - 1 \cdot x + 1 x + 1 \cdot - 1)$

Rewrite $- 1 x$ as $- x$ .

$9 x + 5 = 4 (x^{2} - x + 1 x + 1 \cdot - 1)$

Multiply $x$ by $1$ .

$9 x + 5 = 4 (x^{2} - x + x + 1 \cdot - 1)$

Multiply $- 1$ by $1$ .

$9 x + 5 = 4 (x^{2} - x + x - 1)$

Add $- x$ and $x$ .

$9 x + 5 = 4 (x^{2} + 0 - 1)$

Add $x^{2}$ and $0$ .

$9 x + 5 = 4 (x^{2} - 1)$

Apply the distributive property.

$9 x + 5 = 4 x^{2} + 4 \cdot - 1$

Multiply $4$ by $- 1$ .

$9 x + 5 = 4 x^{2} - 4$

Subtract $4 x^{2}$ from both sides of the equation.

$9 x + 5 - 4 x^{2} = - 4$

Move all terms not containing $x$ to the right side of the equation.

Subtract $5$ from both sides of the equation.

$9 x - 4 x^{2} = - 4 - 5$

Subtract $5$ from $- 4$ .

$9 x - 4 x^{2} = - 9$

Add $9$ to both sides of the equation.

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

Factor the left side of the equation.

Factor $- 1$ out of $9 x - 4 x^{2} + 9$ .

Reorder $9 x$ and $- 4 x^{2}$ .

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

Factor $- 1$ out of $- 4 x^{2}$ .

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

Factor $- 1$ out of $9 x$ .

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

Rewrite $9$ as $- 1 (- 9)$ .

$- (4 x^{2}) - (- 9 x) - 1 \cdot - 9 = 0$

Factor $- 1$ out of $- (4 x^{2}) - (- 9 x)$ .

$- (4 x^{2} - 9 x) - 1 \cdot - 9 = 0$

Factor $- 1$ out of $- (4 x^{2} - 9 x) - 1 (- 9)$ .

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

Factor.

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 $a \cdot c = 4 \cdot - 9 = - 36$ and whose sum is $b = - 9$ .

Factor $- 9$ out of $- 9 x$ .

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

Rewrite $- 9$ as $3$ plus $- 12$

$- (4 x^{2} + (3 - 12) x - 9) = 0$

Apply the distributive property.

$- (4 x^{2} + 3 x - 12 x - 9) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$- ((4 x^{2} + 3 x) - 12 x - 9) = 0$

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

$- (x (4 x + 3) - 3 (4 x + 3)) = 0$

Factor the polynomial by factoring out the greatest common factor, $4 x + 3$ .

$- ((4 x + 3) (x - 3)) = 0$

Remove unnecessary parentheses.

$- (4 x + 3) (x - 3) = 0$

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

$4 x + 3 = 0$

$x - 3 = 0$

Set $4 x + 3$ equal to $0$ and solve for $x$ .

Set $4 x + 3$ equal to $0$ .

$4 x + 3 = 0$

Solve $4 x + 3 = 0$ for $x$ .

Subtract $3$ from both sides of the equation.

$4 x = - 3$

Divide each term in $4 x = - 3$ by $4$ and simplify.

Divide each term in $4 x = - 3$ by $4$ .

$\frac{4 x}{4} = \frac{- 3}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{- 3}{4}$

Divide $x$ by $1$ .

$x = \frac{- 3}{4}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{3}{4}$

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

The final solution is all the values that make $- (4 x + 3) (x - 3) = 0$ true.

$x = - \frac{3}{4}, 3$

Exclude the solutions that do not make $\log_{16} (9 x + 5) - \log_{16} (x^{2} - 1) = \frac{1}{2}$ true.

$x = 3$

Solve for x log base 16 of 9x+5- log base 16 of x^2-1=1/2

Related Questions

Support LCMGCF.com by disabling your adblocker.