Solve the Inequality for x 1/16*(1/8)^x<=(1/4)^x

$\frac{1}{16} \cdot {(\frac{1}{8})}^{x} \leq {(\frac{1}{4})}^{x}$

Simplify $\frac{1}{16} \cdot {(\frac{1}{8})}^{x}$ .

Rewrite.

$0 + 0 + \frac{1}{16} \cdot {(\frac{1}{8})}^{x} \leq {(\frac{1}{4})}^{x}$

Apply the product rule to $\frac{1}{8}$ .

$\frac{1}{16} \cdot \frac{1^{x}}{8^{x}} \leq {(\frac{1}{4})}^{x}$

Combine.

$\frac{1 \cdot 1^{x}}{16 \cdot 8^{x}} \leq {(\frac{1}{4})}^{x}$

Multiply $1$ by $1^{x}$ .

Raise $1$ to the power of $1$ .

$\frac{1^{1} \cdot 1^{x}}{16 \cdot 8^{x}} \leq {(\frac{1}{4})}^{x}$

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

$\frac{1^{1 + x}}{16 \cdot 8^{x}} \leq {(\frac{1}{4})}^{x}$

One to any power is one.

$\frac{1}{16 \cdot 8^{x}} \leq {(\frac{1}{4})}^{x}$

Simplify the denominator.

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

$\frac{1}{2^{4} \cdot 8^{x}} \leq {(\frac{1}{4})}^{x}$

Rewrite $8$ as $2^{3}$ .

$\frac{1}{2^{4} \cdot {(2^{3})}^{x}} \leq {(\frac{1}{4})}^{x}$

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

$\frac{1}{2^{4} \cdot 2^{3 x}} \leq {(\frac{1}{4})}^{x}$

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

$\frac{1}{2^{4 + 3 x}} \leq {(\frac{1}{4})}^{x}$

Simplify ${(\frac{1}{4})}^{x}$ .

Apply the product rule to $\frac{1}{4}$ .

$\frac{1}{2^{4 + 3 x}} \leq \frac{1^{x}}{4^{x}}$

One to any power is one.

$\frac{1}{2^{4 + 3 x}} \leq \frac{1}{4^{x}}$

Subtract $\frac{1}{4^{x}}$ from both sides of the inequality.

$\frac{1}{2^{4 + 3 x}} - \frac{1}{4^{x}} \leq 0$

Move $- \frac{1}{4^{x}}$ to the right side of the equation by adding it to both sides.

$\frac{1}{2^{4 + 3 x}} = \frac{1}{4^{x}}$

Take the log of both sides of the equation.

$\ln (\frac{1}{2^{4 + 3 x}}) = \ln (\frac{1}{4^{x}})$

Rewrite $\ln (\frac{1}{2^{4 + 3 x}})$ as $\ln (1) - \ln (2^{4 + 3 x})$ .

$\ln (1) - \ln (2^{4 + 3 x}) = \ln (\frac{1}{4^{x}})$

Expand $\ln (2^{4 + 3 x})$ by moving $4 + 3 x$ outside the logarithm.

$\ln (1) - ((4 + 3 x) \ln (2)) = \ln (\frac{1}{4^{x}})$

The natural logarithm of $1$ is $0$ .

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

Subtract $(4 + 3 x) \ln (2)$ from $0$ .

$- ((4 + 3 x) \ln (2)) = \ln (\frac{1}{4^{x}})$

Remove parentheses.

$- (4 + 3 x) \ln (2) = \ln (\frac{1}{4^{x}})$

Rewrite $\ln (\frac{1}{4^{x}})$ as $\ln (1) - \ln (4^{x})$ .

$- (4 + 3 x) \ln (2) = \ln (1) - \ln (4^{x})$

Expand $\ln (4^{x})$ by moving $x$ outside the logarithm.

$- (4 + 3 x) \ln (2) = \ln (1) - (x \ln (4))$

The natural logarithm of $1$ is $0$ .

$- (4 + 3 x) \ln (2) = 0 - (x \ln (4))$

Subtract $x \ln (4)$ from $0$ .

$- (4 + 3 x) \ln (2) = - (x \ln (4))$

Remove parentheses.

$- (4 + 3 x) \ln (2) = - x \ln (4)$

Solve the equation for $x$ .

Simplify $- (4 + 3 x) \ln (2)$ .

Start simplifying.

$0 + 0 - (4 + 3 x) \ln (2) = - x \ln (4)$

Simplify by adding zeros.

$- (4 + 3 x) \ln (2) = - x \ln (4)$

Apply the distributive property.

$(- 1 \cdot 4 - (3 x)) \ln (2) = - x \ln (4)$

Multiply.

Multiply $- 1$ by $4$ .

$(- 4 - (3 x)) \ln (2) = - x \ln (4)$

Multiply $3$ by $- 1$ .

$(- 4 - 3 x) \ln (2) = - x \ln (4)$

Apply the distributive property.

$- 4 \ln (2) - 3 x \ln (2) = - x \ln (4)$

Add $x \ln (4)$ to both sides of the equation.

$- 4 \ln (2) - 3 x \ln (2) + x \ln (4) = 0$

Add $4 \ln (2)$ to both sides of the equation.

$- 3 x \ln (2) + x \ln (4) = 4 \ln (2)$

Factor $x$ out of $- 3 x \ln (2) + x \ln (4)$ .

Factor $x$ out of $- 3 x \ln (2)$ .

$x (- 3 \ln (2)) + x \ln (4) = 4 \ln (2)$

Factor $x$ out of $x \ln (4)$ .

$x (- 3 \ln (2)) + x (\ln (4)) = 4 \ln (2)$

Factor $x$ out of $x (- 3 \ln (2)) + x (\ln (4))$ .

$x (- 3 \ln (2) + \ln (4)) = 4 \ln (2)$

Divide each term in $x (- 3 \ln (2) + \ln (4)) = 4 \ln (2)$ by $- 3 \ln (2) + \ln (4)$ and simplify.

Divide each term in $x (- 3 \ln (2) + \ln (4)) = 4 \ln (2)$ by $- 3 \ln (2) + \ln (4)$ .

$\frac{x (- 3 \ln (2) + \ln (4))}{- 3 \ln (2) + \ln (4)} = \frac{4 \ln (2)}{- 3 \ln (2) + \ln (4)}$

Simplify the left side.

Cancel the common factor of $- 3 \ln (2) + \ln (4)$ .

Cancel the common factor.

$\frac{x (- 3 \ln (2) + \ln (4))}{- 3 \ln (2) + \ln (4)} = \frac{4 \ln (2)}{- 3 \ln (2) + \ln (4)}$

Divide $x$ by $1$ .

$x = \frac{4 \ln (2)}{- 3 \ln (2) + \ln (4)}$

Simplify the right side.

Factor $- 1$ out of $- 3 \ln (2)$ .

$x = \frac{4 \ln (2)}{- (3 \ln (2)) + \ln (4)}$

Factor $- 1$ out of $\ln (4)$ .

$x = \frac{4 \ln (2)}{- (3 \ln (2)) - 1 (- \ln (4))}$

Factor $- 1$ out of $- (3 \ln (2)) - 1 (- \ln (4))$ .

$x = \frac{4 \ln (2)}{- (3 \ln (2) - \ln (4))}$

Rewrite negatives.

Rewrite $- (3 \ln (2) - \ln (4))$ as $- 1 (3 \ln (2) - \ln (4))$ .

$x = \frac{4 \ln (2)}{- 1 (3 \ln (2) - \ln (4))}$

Move the negative in front of the fraction.

$x = - \frac{4 \ln (2)}{3 \ln (2) - \ln (4)}$

The solution consists of all of the true intervals.

$x \geq - \frac{4 \ln (2)}{3 \ln (2) - \ln (4)}$

The result can be shown in multiple forms.

Inequality Form:

$x \geq - \frac{4 \ln (2)}{3 \ln (2) - \ln (4)}$

Interval Notation:

$[- \frac{4 \ln (2)}{3 \ln (2) - \ln (4)}, \infty)$

Solve the Inequality for x 1/16*(1/8)^x<=(1/4)^x

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