Find Where Undefined/Discontinuous log base 2 of log base 2 of square root of 4x=1

$\log_{2} (\log_{2} (\sqrt{4 x})) = 1$

Step 1

Subtract $1$ from both sides of the equation.

$\log_{2} (\log_{2} (\sqrt{4 x})) - 1 = 0$

Step 2

Simplify each term.

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

$\log_{2} (\log_{2} (\sqrt{2^{2} x})) - 1 = 0$

Pull terms out from under the radical.

$\log_{2} (\log_{2} (2 \sqrt{x})) - 1 = 0$

Step 3

Set the argument in $\log_{2} (2 \sqrt{x})$ less than or equal to $0$ to find where the expression is undefined.

$2 \sqrt{x} \leq 0$

Step 4

Solve for $x$ .

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

${(2 \sqrt{x})}^{2} \leq 0^{2}$

Simplify each side of the inequality.

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

${(2 x^{\frac{1}{2}})}^{2} \leq 0^{2}$

Simplify the left side.

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

Apply the product rule to $2 x^{\frac{1}{2}}$ .

$2^{2} {(x^{\frac{1}{2}})}^{2} \leq 0^{2}$

Raise $2$ to the power of $2$ .

$4 {(x^{\frac{1}{2}})}^{2} \leq 0^{2}$

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

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

$4 x^{\frac{1}{2} \cdot 2} \leq 0^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$4 x^{\frac{1}{2} \cdot 2} \leq 0^{2}$

Rewrite the expression.

$4 x^{1} \leq 0^{2}$

Simplify.

$4 x \leq 0^{2}$

Simplify the right side.

Raising $0$ to any positive power yields $0$ .

$4 x \leq 0$

Divide each term in $4 x \leq 0$ by $4$ and simplify.

Divide each term in $4 x \leq 0$ by $4$ .

$\frac{4 x}{4} \leq \frac{0}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} \leq \frac{0}{4}$

Divide $x$ by $1$ .

$x \leq \frac{0}{4}$

Simplify the right side.

Divide $0$ by $4$ .

$x \leq 0$

Step 5

Set the argument in $\log_{2} (\log_{2} (2 \sqrt{x}))$ less than or equal to $0$ to find where the expression is undefined.

$\log_{2} (2 \sqrt{x}) \leq 0$

Step 6

Solve for $x$ .

Convert the inequality to an equality.

$\log_{2} (2 \sqrt{x}) = 0$

Solve the equation.

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

$2^{0} = 2 \sqrt{x}$

Solve for $x$ .

Rewrite the equation as $2 \sqrt{x} = 2^{0}$ .

$2 \sqrt{x} = 2^{0}$

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

${(2 \sqrt{x})}^{2} = {(2^{0})}^{2}$

Simplify each side of the equation.

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

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

Simplify the left side.

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

Apply the product rule to $2 x^{\frac{1}{2}}$ .

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

Raise $2$ to the power of $2$ .

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

$4 x^{1} = {(2^{0})}^{2}$

Simplify.

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

Simplify the right side.

Simplify ${(2^{0})}^{2}$ .

Multiply the exponents in ${(2^{0})}^{2}$ .

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

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

Multiply $0$ by $2$ .

$4 x = 2^{0}$

Anything raised to $0$ is $1$ .

$4 x = 1$

Divide each term in $4 x = 1$ by $4$ and simplify.

Divide each term in $4 x = 1$ by $4$ .

$\frac{4 x}{4} = \frac{1}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{1}{4}$

Divide $x$ by $1$ .

$x = \frac{1}{4}$

Find the domain of $\log_{2} (2 \sqrt{x})$ .

Set the argument in $\log_{2} (2 \sqrt{x})$ greater than $0$ to find where the expression is defined.

$2 \sqrt{x} > 0$

Solve for $x$ .

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

${(2 \sqrt{x})}^{2} > 0^{2}$

Simplify each side of the inequality.

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

${(2 x^{\frac{1}{2}})}^{2} > 0^{2}$

Simplify the left side.

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

Apply the product rule to $2 x^{\frac{1}{2}}$ .

$2^{2} {(x^{\frac{1}{2}})}^{2} > 0^{2}$

Raise $2$ to the power of $2$ .

$4 {(x^{\frac{1}{2}})}^{2} > 0^{2}$

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

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

$4 x^{\frac{1}{2} \cdot 2} > 0^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$4 x^{\frac{1}{2} \cdot 2} > 0^{2}$

Rewrite the expression.

$4 x^{1} > 0^{2}$

Simplify.

$4 x > 0^{2}$

Simplify the right side.

Raising $0$ to any positive power yields $0$ .

$4 x > 0$

Divide each term in $4 x > 0$ by $4$ and simplify.

Divide each term in $4 x > 0$ by $4$ .

$\frac{4 x}{4} > \frac{0}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} > \frac{0}{4}$

Divide $x$ by $1$ .

$x > \frac{0}{4}$

Simplify the right side.

Divide $0$ by $4$ .

$x > 0$

Find the domain of $2 \sqrt{x}$ .

Set the radicand in $\sqrt{x}$ greater than or equal to $0$ to find where the expression is defined.

$x \geq 0$

The domain is all values of $x$ that make the expression defined.

$[0, \infty)$

The solution consists of all of the true intervals.

$x > 0$

Set the radicand in $\sqrt{x}$ greater than or equal to $0$ to find where the expression is defined.

$x \geq 0$

The domain is all values of $x$ that make the expression defined.

$(0, \infty)$

Use each root to create test intervals.

$x < 0$

$0 < x < \frac{1}{4}$

$x > \frac{1}{4}$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < 0$ to see if it makes the inequality true.

Choose a value on the interval $x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Replace $x$ with $- 2$ in the original inequality.

$\log_{2} (2 \sqrt{- 2}) \leq 0$

The left side is not equal to the right side, which means that the given statement is false.

False

Test a value on the interval $0 < x < \frac{1}{4}$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < \frac{1}{4}$ and see if this value makes the original inequality true.

$x = 0.12$

Replace $x$ with $0.12$ in the original inequality.

$\log_{2} (2 \sqrt{0.12}) \leq 0$

The left side $- 0.52944684$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > \frac{1}{4}$ to see if it makes the inequality true.

Choose a value on the interval $x > \frac{1}{4}$ and see if this value makes the original inequality true.

$x = 3$

Replace $x$ with $3$ in the original inequality.

$\log_{2} (2 \sqrt{3}) \leq 0$

The left side $1.79248125$ is greater than the right side $0$ , which means that the given statement is false.

False

Compare the intervals to determine which ones satisfy the original inequality.

$x < 0$ False

$0 < x < \frac{1}{4}$ True

$x > \frac{1}{4}$ False

$x < 0$ False

$0 < x < \frac{1}{4}$ True

$x > \frac{1}{4}$ False

The solution consists of all of the true intervals.

$0 < x \leq \frac{1}{4}$

Step 7

Set the radicand in $\sqrt{x}$ less than $0$ to find where the expression is undefined.

$x < 0$

Step 8

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$x \leq \frac{1}{4}$

$(- \infty, \frac{1}{4}]$

Step 9

Find Where Undefined/Discontinuous log base 2 of log base 2 of square root of 4x=1

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