Solve for x 4 log base 2 of x+ log base 2 of 5 = log base 2 of 405

$4 \log_{2} (x) + \log_{2} (5) = \log_{2} (405)$

Step 1

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

$4 \log_{2} (x) + \log_{2} (5) - \log_{2} (405) = 0$

Step 2

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

$4 \log_{2} (x) + \log_{2} (\frac{5}{405}) = 0$

Step 3

Cancel the common factor of $5$ and $405$ .

Factor $5$ out of $5$ .

$4 \log_{2} (x) + \log_{2} (\frac{5 (1)}{405}) = 0$

Cancel the common factors.

Factor $5$ out of $405$ .

$4 \log_{2} (x) + \log_{2} (\frac{5 \cdot 1}{5 \cdot 81}) = 0$

Cancel the common factor.

$4 \log_{2} (x) + \log_{2} (\frac{5 \cdot 1}{5 \cdot 81}) = 0$

Rewrite the expression.

$4 \log_{2} (x) + \log_{2} (\frac{1}{81}) = 0$

Step 4

Simplify the left side.

Simplify $4 \log_{2} (x) + \log_{2} (\frac{1}{81})$ .

Simplify $4 \log_{2} (x)$ by moving $4$ inside the logarithm.

$\log_{2} (x^{4}) + \log_{2} (\frac{1}{81}) = 0$

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\log_{2} (x^{4} \frac{1}{81}) = 0$

Combine $x^{4}$ and $\frac{1}{81}$ .

$\log_{2} (\frac{x^{4}}{81}) = 0$

Step 5

Rewrite $\log_{2} (\frac{x^{4}}{81}) = 0$ 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 $b^{y} = x$ .

$2^{0} = \frac{x^{4}}{81}$

Step 6

Solve for $x$ .

Rewrite the equation as $\frac{x^{4}}{81} = 2^{0}$ .

$\frac{x^{4}}{81} = 2^{0}$

Multiply both sides of the equation by $81$ .

$81 \frac{x^{4}}{81} = 81 \cdot 2^{0}$

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $81$ .

Cancel the common factor.

$81 \frac{x^{4}}{81} = 81 \cdot 2^{0}$

Rewrite the expression.

$x^{4} = 81 \cdot 2^{0}$

Simplify the right side.

Simplify $81 \cdot 2^{0}$ .

Anything raised to $0$ is $1$ .

$x^{4} = 81 \cdot 1$

Multiply $81$ by $1$ .

$x^{4} = 81$

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

$x = \pm \sqrt[4]{81}$

Simplify $\pm \sqrt[4]{81}$ .

Rewrite $81$ as $3^{4}$ .

$x = \pm \sqrt[4]{3^{4}}$

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

$x = \pm 3$

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.

$x = 3$

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

$x = - 3$

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

$x = 3, - 3$

Step 7

Exclude the solutions that do not make $4 \log_{2} (x) + \log_{2} (5) = \log_{2} (405)$ true.

$x = 3$

Solve for x 4 log base 2 of x+ log base 2 of 5 = log base 2 of 405

Related Questions

Support LCMGCF.com by disabling your adblocker.