Solve for x log base 9 of square root of 10x+5-1/2 = log base 9 of square root of x+1

$\log_{9} (\sqrt{10 x + 5}) - \frac{1}{2} = \log_{9} (\sqrt{x + 1})$

Step 1

Move all the terms containing a logarithm to the left side of the equation.

$\log_{9} (\sqrt{10 x + 5}) - \log_{9} (\sqrt{x + 1}) = \frac{1}{2}$

Step 2

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

$\log_{9} (\frac{\sqrt{10 x + 5}}{\sqrt{x + 1}}) = \frac{1}{2}$

Step 3

Factor $5$ out of $10 x + 5$ .

Factor $5$ out of $10 x$ .

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

Factor $5$ out of $5$ .

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

Factor $5$ out of $5 (2 x) + 5 (1)$ .

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

Step 4

Multiply $\frac{\sqrt{5 (2 x + 1)}}{\sqrt{x + 1}}$ by $\frac{\sqrt{x + 1}}{\sqrt{x + 1}}$ .

$\log_{9} (\frac{\sqrt{5 (2 x + 1)}}{\sqrt{x + 1}} \cdot \frac{\sqrt{x + 1}}{\sqrt{x + 1}}) = \frac{1}{2}$

Step 5

Combine and simplify the denominator.

Multiply $\frac{\sqrt{5 (2 x + 1)}}{\sqrt{x + 1}}$ by $\frac{\sqrt{x + 1}}{\sqrt{x + 1}}$ .

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

Raise $\sqrt{x + 1}$ to the power of $1$ .

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

Raise $\sqrt{x + 1}$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $1$ and $1$ .

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

Rewrite ${\sqrt{x + 1}}^{2}$ as $x + 1$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x + 1}$ as ${(x + 1)}^{\frac{1}{2}}$ .

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

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

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

Combine $\frac{1}{2}$ and $2$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Step 6

Combine using the product rule for radicals.

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

Step 7

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

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

Step 8

Cross multiply to remove the fraction.

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

Step 9

Simplify $9^{\frac{1}{2}} (x + 1)$ .

Simplify the expression.

Rewrite $9$ as $3^{2}$ .

$\sqrt{5 (2 x + 1) (x + 1)} = {(3^{2})}^{\frac{1}{2}} (x + 1)$

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

$\sqrt{5 (2 x + 1) (x + 1)} = 3^{2 (\frac{1}{2})} (x + 1)$

Cancel the common factor of $2$ .

Cancel the common factor.

$\sqrt{5 (2 x + 1) (x + 1)} = 3^{2 (\frac{1}{2})} (x + 1)$

Rewrite the expression.

$\sqrt{5 (2 x + 1) (x + 1)} = 3^{1} (x + 1)$

Evaluate the exponent.

$\sqrt{5 (2 x + 1) (x + 1)} = 3 (x + 1)$

Apply the distributive property.

$\sqrt{5 (2 x + 1) (x + 1)} = 3 x + 3 \cdot 1$

Multiply $3$ by $1$ .

$\sqrt{5 (2 x + 1) (x + 1)} = 3 x + 3$

Step 10

Subtract $3 x$ from both sides of the equation.

$\sqrt{5 (2 x + 1) (x + 1)} - 3 x = 3$

Step 11

Add $3 x$ to both sides of the equation.

$\sqrt{5 (2 x + 1) (x + 1)} = 3 + 3 x$

Step 12

To remove the radical on the left side of the equation, square both sides of the equation.

${\sqrt{5 (2 x + 1) (x + 1)}}^{2} = {(3 + 3 x)}^{2}$

Step 13

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{5 (2 x + 1) (x + 1)}$ as ${(5 (2 x + 1) (x + 1))}^{\frac{1}{2}}$ .

${({(5 (2 x + 1) (x + 1))}^{\frac{1}{2}})}^{2} = {(3 + 3 x)}^{2}$

Simplify the left side.

Simplify ${({(5 (2 x + 1) (x + 1))}^{\frac{1}{2}})}^{2}$ .

Simplify by multiplying through.

Multiply the exponents in ${({(5 (2 x + 1) (x + 1))}^{\frac{1}{2}})}^{2}$ .

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

${(5 (2 x + 1) (x + 1))}^{\frac{1}{2} \cdot 2} = {(3 + 3 x)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(5 (2 x + 1) (x + 1))}^{\frac{1}{2} \cdot 2} = {(3 + 3 x)}^{2}$

Rewrite the expression.

${(5 (2 x + 1) (x + 1))}^{1} = {(3 + 3 x)}^{2}$

Apply the distributive property.

${((5 (2 x) + 5 \cdot 1) (x + 1))}^{1} = {(3 + 3 x)}^{2}$

Multiply.

Multiply $2$ by $5$ .

${((10 x + 5 \cdot 1) (x + 1))}^{1} = {(3 + 3 x)}^{2}$

Multiply $5$ by $1$ .

${((10 x + 5) (x + 1))}^{1} = {(3 + 3 x)}^{2}$

Expand $(10 x + 5) (x + 1)$ using the FOIL Method.

Apply the distributive property.

${(10 x (x + 1) + 5 (x + 1))}^{1} = {(3 + 3 x)}^{2}$

Apply the distributive property.

${(10 x \cdot x + 10 x \cdot 1 + 5 (x + 1))}^{1} = {(3 + 3 x)}^{2}$

Apply the distributive property.

${(10 x \cdot x + 10 x \cdot 1 + 5 x + 5 \cdot 1)}^{1} = {(3 + 3 x)}^{2}$

Simplify and combine like terms.

Simplify each term.

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

Move $x$ .

${(10 (x \cdot x) + 10 x \cdot 1 + 5 x + 5 \cdot 1)}^{1} = {(3 + 3 x)}^{2}$

Multiply $x$ by $x$ .

${(10 x^{2} + 10 x \cdot 1 + 5 x + 5 \cdot 1)}^{1} = {(3 + 3 x)}^{2}$

Multiply $10$ by $1$ .

${(10 x^{2} + 10 x + 5 x + 5 \cdot 1)}^{1} = {(3 + 3 x)}^{2}$

Multiply $5$ by $1$ .

${(10 x^{2} + 10 x + 5 x + 5)}^{1} = {(3 + 3 x)}^{2}$

Add $10 x$ and $5 x$ .

${(10 x^{2} + 15 x + 5)}^{1} = {(3 + 3 x)}^{2}$

Simplify.

$10 x^{2} + 15 x + 5 = {(3 + 3 x)}^{2}$

Simplify the right side.

Simplify ${(3 + 3 x)}^{2}$ .

Rewrite ${(3 + 3 x)}^{2}$ as $(3 + 3 x) (3 + 3 x)$ .

$10 x^{2} + 15 x + 5 = (3 + 3 x) (3 + 3 x)$

Expand $(3 + 3 x) (3 + 3 x)$ using the FOIL Method.

Apply the distributive property.

$10 x^{2} + 15 x + 5 = 3 (3 + 3 x) + 3 x (3 + 3 x)$

Apply the distributive property.

$10 x^{2} + 15 x + 5 = 3 \cdot 3 + 3 (3 x) + 3 x (3 + 3 x)$

Apply the distributive property.

$10 x^{2} + 15 x + 5 = 3 \cdot 3 + 3 (3 x) + 3 x \cdot 3 + 3 x (3 x)$

Simplify and combine like terms.

Simplify each term.

Multiply $3$ by $3$ .

$10 x^{2} + 15 x + 5 = 9 + 3 (3 x) + 3 x \cdot 3 + 3 x (3 x)$

Multiply $3$ by $3$ .

$10 x^{2} + 15 x + 5 = 9 + 9 x + 3 x \cdot 3 + 3 x (3 x)$

Multiply $3$ by $3$ .

$10 x^{2} + 15 x + 5 = 9 + 9 x + 9 x + 3 x (3 x)$

Rewrite using the commutative property of multiplication.

$10 x^{2} + 15 x + 5 = 9 + 9 x + 9 x + 3 \cdot 3 x \cdot x$

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

Move $x$ .

$10 x^{2} + 15 x + 5 = 9 + 9 x + 9 x + 3 \cdot 3 (x \cdot x)$

Multiply $x$ by $x$ .

$10 x^{2} + 15 x + 5 = 9 + 9 x + 9 x + 3 \cdot 3 x^{2}$

Multiply $3$ by $3$ .

$10 x^{2} + 15 x + 5 = 9 + 9 x + 9 x + 9 x^{2}$

Add $9 x$ and $9 x$ .

$10 x^{2} + 15 x + 5 = 9 + 18 x + 9 x^{2}$

Step 14

Solve for $x$ .

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

Subtract $18 x$ from both sides of the equation.

$10 x^{2} + 15 x + 5 - 18 x = 9 + 9 x^{2}$

Subtract $9 x^{2}$ from both sides of the equation.

$10 x^{2} + 15 x + 5 - 18 x - 9 x^{2} = 9$

Subtract $9 x^{2}$ from $10 x^{2}$ .

$x^{2} + 15 x + 5 - 18 x = 9$

Subtract $18 x$ from $15 x$ .

$x^{2} - 3 x + 5 = 9$

Subtract $9$ from both sides of the equation.

$x^{2} - 3 x + 5 - 9 = 0$

Subtract $9$ from $5$ .

$x^{2} - 3 x - 4 = 0$

Factor $x^{2} - 3 x - 4$ 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 $- 4$ and whose sum is $- 3$ .

$- 4, 1$

Write the factored form using these integers.

$(x - 4) (x + 1) = 0$

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

$x - 4 = 0$

$x + 1 = 0$

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

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

The final solution is all the values that make $(x - 4) (x + 1) = 0$ true.

$x = 4, - 1$

Step 15

Exclude the solutions that do not make $\log_{9} (\sqrt{10 x + 5}) - \frac{1}{2} = \log_{9} (\sqrt{x + 1})$ true.

$x = 4$

Solve for x log base 9 of square root of 10x+5-1/2 = log base 9 of square root of x+1

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