Solve for q 1/4* natural log of 16q^8- natural log of 3 = natural log of 24

$\frac{1}{4} \cdot \ln (16 q^{8}) - \ln (3) = \ln (24)$

Simplify the left side.

Combine $\frac{1}{4}$ and $\ln (16 q^{8})$ .

$\frac{\ln (16 q^{8})}{4} - \ln (3) = \ln (24)$

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

$\frac{\ln (16 q^{8})}{4} - \ln (3) - \ln (24) = 0$

Simplify the left side.

Simplify $\frac{\ln (16 q^{8})}{4} - \ln (3) - \ln (24)$ .

Simplify each term.

Rewrite $\frac{\ln (16 q^{8})}{4}$ as $\frac{1}{4} \ln (16 q^{8})$ .

$\frac{1}{4} \ln (16 q^{8}) - \ln (3) - \ln (24) = 0$

Simplify $\frac{1}{4} \ln (16 q^{8})$ by moving $\frac{1}{4}$ inside the logarithm.

$\ln ({(16 q^{8})}^{\frac{1}{4}}) - \ln (3) - \ln (24) = 0$

Apply the product rule to $16 q^{8}$ .

$\ln (16^{\frac{1}{4}} {(q^{8})}^{\frac{1}{4}}) - \ln (3) - \ln (24) = 0$

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

$\ln ({(2^{4})}^{\frac{1}{4}} {(q^{8})}^{\frac{1}{4}}) - \ln (3) - \ln (24) = 0$

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

$\ln (2^{4 (\frac{1}{4})} {(q^{8})}^{\frac{1}{4}}) - \ln (3) - \ln (24) = 0$

Cancel the common factor of $4$ .

Cancel the common factor.

$\ln (2^{4 (\frac{1}{4})} {(q^{8})}^{\frac{1}{4}}) - \ln (3) - \ln (24) = 0$

Rewrite the expression.

$\ln (2^{1} {(q^{8})}^{\frac{1}{4}}) - \ln (3) - \ln (24) = 0$

Evaluate the exponent.

$\ln (2 {(q^{8})}^{\frac{1}{4}}) - \ln (3) - \ln (24) = 0$

Multiply the exponents in ${(q^{8})}^{\frac{1}{4}}$ .

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

$\ln (2 q^{8 (\frac{1}{4})}) - \ln (3) - \ln (24) = 0$

Cancel the common factor of $4$ .

Factor $4$ out of $8$ .

$\ln (2 q^{4 (2) \frac{1}{4}}) - \ln (3) - \ln (24) = 0$

Cancel the common factor.

$\ln (2 q^{4 \cdot 2 \frac{1}{4}}) - \ln (3) - \ln (24) = 0$

Rewrite the expression.

$\ln (2 q^{2}) - \ln (3) - \ln (24) = 0$

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

$\ln (\frac{2 q^{2}}{3}) - \ln (24) = 0$

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

$\ln (\frac{\frac{2 q^{2}}{3}}{24}) = 0$

Multiply the numerator by the reciprocal of the denominator.

$\ln (\frac{2 q^{2}}{3} \cdot \frac{1}{24}) = 0$

Combine.

$\ln (\frac{2 q^{2} \cdot 1}{3 \cdot 24}) = 0$

Cancel the common factor of $2$ and $24$ .

Factor $2$ out of $2 q^{2} \cdot 1$ .

$\ln (\frac{2 (q^{2} \cdot 1)}{3 \cdot 24}) = 0$

Cancel the common factors.

Factor $2$ out of $3 \cdot 24$ .

$\ln (\frac{2 (q^{2} \cdot 1)}{2 (3 \cdot 12)}) = 0$

Cancel the common factor.

$\ln (\frac{2 (q^{2} \cdot 1)}{2 (3 \cdot 12)}) = 0$

Rewrite the expression.

$\ln (\frac{q^{2} \cdot 1}{3 \cdot 12}) = 0$

Multiply.

Multiply $q^{2}$ by $1$ .

$\ln (\frac{q^{2}}{3 \cdot 12}) = 0$

Multiply $3$ by $12$ .

$\ln (\frac{q^{2}}{36}) = 0$

To solve for $q$ , rewrite the equation using properties of logarithms.

$e^{\ln (\frac{q^{2}}{36})} = e^{0}$

Rewrite $\ln (\frac{q^{2}}{36}) = 0$ 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 $b^{y} = x$ .

$e^{0} = \frac{q^{2}}{36}$

Solve for $q$ .

Rewrite the equation as $\frac{q^{2}}{36} = e^{0}$ .

$\frac{q^{2}}{36} = e^{0}$

Multiply both sides of the equation by $36$ .

$36 \frac{q^{2}}{36} = 36 e^{0}$

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $36$ .

Cancel the common factor.

$36 \frac{q^{2}}{36} = 36 e^{0}$

Rewrite the expression.

$q^{2} = 36 e^{0}$

Simplify the right side.

Simplify $36 e^{0}$ .

Anything raised to $0$ is $1$ .

$q^{2} = 36 \cdot 1$

Multiply $36$ by $1$ .

$q^{2} = 36$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$q = \pm \sqrt{36}$

Simplify $\pm \sqrt{36}$ .

Rewrite $36$ as $6^{2}$ .

$q = \pm \sqrt{6^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$q = \pm 6$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$q = 6$

Next, use the negative value of the $\pm$ to find the second solution.

$q = - 6$

The complete solution is the result of both the positive and negative portions of the solution.

$q = 6, - 6$

Solve for q 1/4* natural log of 16q^8- natural log of 3 = natural log of 24

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