Solve for k log base 5 of 3k+12=3/4*( log base 5 of 405- log base 5 of 5)

$\log_{5} (3 k + 12) = \frac{3}{4} \cdot (\log_{5} (405) - \log_{5} (5))$

Simplify the right side.

Simplify $\frac{3}{4} \cdot (\log_{5} (405) - \log_{5} (5))$ .

Apply the distributive property.

$\log_{5} (3 k + 12) = \frac{3}{4} \log_{5} (405) + \frac{3}{4} (- \log_{5} (5))$

Combine $\frac{3}{4}$ and $\log_{5} (405)$ .

$\log_{5} (3 k + 12) = \frac{3 \log_{5} (405)}{4} + \frac{3}{4} (- \log_{5} (5))$

Combine $\frac{3}{4}$ and $\log_{5} (5)$ .

$\log_{5} (3 k + 12) = \frac{3 \log_{5} (405)}{4} - \frac{3 \log_{5} (5)}{4}$

Multiply each term in $\log_{5} (3 k + 12) = \frac{3 \log_{5} (405)}{4} - \frac{3 \log_{5} (5)}{4}$ by $4$ to eliminate the fractions.

Multiply each term in $\log_{5} (3 k + 12) = \frac{3 \log_{5} (405)}{4} - \frac{3 \log_{5} (5)}{4}$ by $4$ .

$\log_{5} (3 k + 12) \cdot 4 = \frac{3 \log_{5} (405)}{4} \cdot 4 - \frac{3 \log_{5} (5)}{4} \cdot 4$

Simplify the left side.

Move $4$ to the left of $\log_{5} (3 k + 12)$ .

$4 \log_{5} (3 k + 12) = \frac{3 \log_{5} (405)}{4} \cdot 4 - \frac{3 \log_{5} (5)}{4} \cdot 4$

Simplify the right side.

Simplify each term.

Cancel the common factor of $4$ .

Cancel the common factor.

$4 \log_{5} (3 k + 12) = \frac{3 \log_{5} (405)}{4} \cdot 4 - \frac{3 \log_{5} (5)}{4} \cdot 4$

Rewrite the expression.

$4 \log_{5} (3 k + 12) = 3 \log_{5} (405) - \frac{3 \log_{5} (5)}{4} \cdot 4$

Cancel the common factor of $4$ .

Move the leading negative in $- \frac{3 \log_{5} (5)}{4}$ into the numerator.

$4 \log_{5} (3 k + 12) = 3 \log_{5} (405) + \frac{- 3 \log_{5} (5)}{4} \cdot 4$

Cancel the common factor.

$4 \log_{5} (3 k + 12) = 3 \log_{5} (405) + \frac{- 3 \log_{5} (5)}{4} \cdot 4$

Rewrite the expression.

$4 \log_{5} (3 k + 12) = 3 \log_{5} (405) - 3 \log_{5} (5)$

Simplify the left side.

Simplify $4 \log_{5} (3 k + 12)$ by moving $4$ inside the logarithm.

$\log_{5} ({(3 k + 12)}^{4}) = 3 \log_{5} (405) - 3 \log_{5} (5)$

Simplify the right side.

Simplify each term.

Simplify $3 \log_{5} (405)$ by moving $3$ inside the logarithm.

$\log_{5} ({(3 k + 12)}^{4}) = \log_{5} (405^{3}) - 3 \log_{5} (5)$

Raise $405$ to the power of $3$ .

$\log_{5} ({(3 k + 12)}^{4}) = \log_{5} (66430125) - 3 \log_{5} (5)$

Logarithm base $5$ of $5$ is $1$ .

$\log_{5} ({(3 k + 12)}^{4}) = \log_{5} (66430125) - 3 \cdot 1$

Multiply $- 3$ by $1$ .

$\log_{5} ({(3 k + 12)}^{4}) = \log_{5} (66430125) - 3$

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

$\log_{5} ({(3 k + 12)}^{4}) - \log_{5} (66430125) = - 3$

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

$\log_{5} (\frac{{(3 k + 12)}^{4}}{66430125}) = - 3$

Simplify the numerator.

Factor $3$ out of $3 k + 12$ .

Factor $3$ out of $3 k$ .

$\log_{5} (\frac{{(3 (k) + 12)}^{4}}{66430125}) = - 3$

Factor $3$ out of $12$ .

$\log_{5} (\frac{{(3 k + 3 \cdot 4)}^{4}}{66430125}) = - 3$

Factor $3$ out of $3 k + 3 \cdot 4$ .

$\log_{5} (\frac{{(3 (k + 4))}^{4}}{66430125}) = - 3$

Apply the product rule to $3 (k + 4)$ .

$\log_{5} (\frac{3^{4} {(k + 4)}^{4}}{66430125}) = - 3$

Raise $3$ to the power of $4$ .

$\log_{5} (\frac{81 {(k + 4)}^{4}}{66430125}) = - 3$

Cancel the common factor of $81$ and $66430125$ .

Factor $81$ out of $81 {(k + 4)}^{4}$ .

$\log_{5} (\frac{81 ({(k + 4)}^{4})}{66430125}) = - 3$

Cancel the common factors.

Factor $81$ out of $66430125$ .

$\log_{5} (\frac{81 {(k + 4)}^{4}}{81 \cdot 820125}) = - 3$

Cancel the common factor.

$\log_{5} (\frac{81 {(k + 4)}^{4}}{81 \cdot 820125}) = - 3$

Rewrite the expression.

$\log_{5} (\frac{{(k + 4)}^{4}}{820125}) = - 3$

Rewrite $\log_{5} (\frac{{(k + 4)}^{4}}{820125}) = - 3$ 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 $b^{y} = x$ .

$5^{- 3} = \frac{{(k + 4)}^{4}}{820125}$

Solve for $k$ .

Rewrite the equation as $\frac{{(k + 4)}^{4}}{820125} = 5^{- 3}$ .

$\frac{{(k + 4)}^{4}}{820125} = 5^{- 3}$

Multiply both sides of the equation by $820125$ .

$820125 \frac{{(k + 4)}^{4}}{820125} = 820125 \cdot 5^{- 3}$

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $820125$ .

Cancel the common factor.

$820125 \frac{{(k + 4)}^{4}}{820125} = 820125 \cdot 5^{- 3}$

Rewrite the expression.

${(k + 4)}^{4} = 820125 \cdot 5^{- 3}$

Simplify the right side.

Simplify $820125 \cdot 5^{- 3}$ .

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

${(k + 4)}^{4} = 820125 \cdot \frac{1}{5^{3}}$

Raise $5$ to the power of $3$ .

${(k + 4)}^{4} = 820125 \cdot \frac{1}{125}$

Cancel the common factor of $125$ .

Factor $125$ out of $820125$ .

${(k + 4)}^{4} = 125 (6561) \cdot \frac{1}{125}$

Cancel the common factor.

${(k + 4)}^{4} = 125 \cdot 6561 \cdot \frac{1}{125}$

Rewrite the expression.

${(k + 4)}^{4} = 6561$

Take the 4th root of both sides of the equation to eliminate the exponent on the left side.

$k + 4 = \pm \sqrt[4]{6561}$

Simplify $\pm \sqrt[4]{6561}$ .

Rewrite $6561$ as $9^{4}$ .

$k + 4 = \pm \sqrt[4]{9^{4}}$

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

$k + 4 = \pm 9$

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.

$k + 4 = 9$

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

Subtract $4$ from both sides of the equation.

$k = 9 - 4$

Subtract $4$ from $9$ .

$k = 5$

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

$k + 4 = - 9$

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

Subtract $4$ from both sides of the equation.

$k = - 9 - 4$

Subtract $4$ from $- 9$ .

$k = - 13$

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

$k = 5, - 13$

Exclude the solutions that do not make $\log_{5} (3 k + 12) = \frac{3}{4} \cdot (\log_{5} (405) - \log_{5} (5))$ true.

$k = 5$

Solve for k log base 5 of 3k+12=3/4*( log base 5 of 405- log base 5 of 5)

Related Questions

Support LCMGCF.com by disabling your adblocker.