Solve the Inequality for x (1/3)^(2x+1)> square root of 27/(3^(x-1))

${(\frac{1}{3})}^{2 x + 1} > \sqrt{\frac{27}{3^{x - 1}}}$

Step 1

Since the radical is on the right side of the equation, switch the sides so it is on the left side of the equation.

$\sqrt{\frac{27}{3^{x - 1}}} < {(\frac{1}{3})}^{2 x + 1}$

Step 2

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

${\sqrt{\frac{27}{3^{x - 1}}}}^{2} < {({(\frac{1}{3})}^{2 x + 1})}^{2}$

Step 3

Simplify each side of the inequality.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{\frac{27}{3^{x - 1}}}$ as ${(\frac{27}{3^{x - 1}})}^{\frac{1}{2}}$ .

${({(\frac{27}{3^{x - 1}})}^{\frac{1}{2}})}^{2} < {({(\frac{1}{3})}^{2 x + 1})}^{2}$

Simplify the left side.

Simplify ${({(\frac{27}{3^{x - 1}})}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(\frac{27}{3^{x - 1}})}^{\frac{1}{2}})}^{2}$ .

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

${(\frac{27}{3^{x - 1}})}^{\frac{1}{2} \cdot 2} < {({(\frac{1}{3})}^{2 x + 1})}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(\frac{27}{3^{x - 1}})}^{\frac{1}{2} \cdot 2} < {({(\frac{1}{3})}^{2 x + 1})}^{2}$

Rewrite the expression.

${(\frac{27}{3^{x - 1}})}^{1} < {({(\frac{1}{3})}^{2 x + 1})}^{2}$

Simplify.

$\frac{27}{3^{x - 1}} < {({(\frac{1}{3})}^{2 x + 1})}^{2}$

Simplify the right side.

Simplify ${({(\frac{1}{3})}^{2 x + 1})}^{2}$ .

Apply the product rule to $\frac{1}{3}$ .

$\frac{27}{3^{x - 1}} < {(\frac{1^{2 x + 1}}{3^{2 x + 1}})}^{2}$

One to any power is one.

$\frac{27}{3^{x - 1}} < {(\frac{1}{3^{2 x + 1}})}^{2}$

Apply the product rule to $\frac{1}{3^{2 x + 1}}$ .

$\frac{27}{3^{x - 1}} < \frac{1^{2}}{{(3^{2 x + 1})}^{2}}$

One to any power is one.

$\frac{27}{3^{x - 1}} < \frac{1}{{(3^{2 x + 1})}^{2}}$

Multiply the exponents in ${(3^{2 x + 1})}^{2}$ .

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

$\frac{27}{3^{x - 1}} < \frac{1}{3^{(2 x + 1) \cdot 2}}$

Apply the distributive property.

$\frac{27}{3^{x - 1}} < \frac{1}{3^{2 x \cdot 2 + 1 \cdot 2}}$

Multiply $2$ by $2$ .

$\frac{27}{3^{x - 1}} < \frac{1}{3^{4 x + 1 \cdot 2}}$

Multiply $2$ by $1$ .

$\frac{27}{3^{x - 1}} < \frac{1}{3^{4 x + 2}}$

Step 4

Solve for $x$ .

Take the log of both sides of the inequality.

$\ln (\frac{27}{3^{x - 1}}) < \ln (\frac{1}{3^{4 x + 2}})$

Rewrite $\ln (\frac{27}{3^{x - 1}})$ as $\ln (27) - \ln (3^{x - 1})$ .

$\ln (27) - \ln (3^{x - 1}) < \ln (\frac{1}{3^{4 x + 2}})$

Expand $\ln (3^{x - 1})$ by moving $x - 1$ outside the logarithm.

$\ln (27) - ((x - 1) \ln (3)) < \ln (\frac{1}{3^{4 x + 2}})$

Remove parentheses.

$\ln (27) - (x - 1) \ln (3) < \ln (\frac{1}{3^{4 x + 2}})$

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

$\ln (27) - (x - 1) \ln (3) < \ln (1) - \ln (3^{4 x + 2})$

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

$\ln (27) - (x - 1) \ln (3) < \ln (1) - ((4 x + 2) \ln (3))$

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

$\ln (27) - (x - 1) \ln (3) < 0 - ((4 x + 2) \ln (3))$

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

$\ln (27) - (x - 1) \ln (3) < - ((4 x + 2) \ln (3))$

Remove parentheses.

$\ln (27) - (x - 1) \ln (3) < - (4 x + 2) \ln (3)$

Solve the inequality for $x$ .

Simplify the left side.

Simplify each term.

Apply the distributive property.

$\ln (27) + (- x + 1) \ln (3) = - (4 x + 2) \ln (3)$

Multiply $- 1$ by $- 1$ .

$\ln (27) + (- x + 1) \ln (3) = - (4 x + 2) \ln (3)$

Apply the distributive property.

$\ln (27) - x \ln (3) + 1 \ln (3) = - (4 x + 2) \ln (3)$

Multiply $\ln (3)$ by $1$ .

$\ln (27) - x \ln (3) + \ln (3) = - (4 x + 2) \ln (3)$

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

$- x \ln (3) + \ln (27 \cdot 3) = - (4 x + 2) \ln (3)$

Multiply $27$ by $3$ .

$- x \ln (3) + \ln (81) = - (4 x + 2) \ln (3)$

Simplify the right side.

Apply the distributive property.

$- x \ln (3) + \ln (81) = (- (4 x) - 1 \cdot 2) \ln (3)$

Multiply.

Multiply $4$ by $- 1$ .

$- x \ln (3) + \ln (81) = (- 4 x - 1 \cdot 2) \ln (3)$

Multiply $- 1$ by $2$ .

$- x \ln (3) + \ln (81) = (- 4 x - 2) \ln (3)$

Apply the distributive property.

$- x \ln (3) + \ln (81) = - 4 x \ln (3) - 2 \ln (3)$

Simplify the right side.

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

Simplify each term.

Simplify $- 4 \ln (3)$ by moving $4$ inside the logarithm.

$- x \ln (3) + \ln (81) = x (- \ln (3^{4})) - 2 \ln (3)$

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

$- x \ln (3) + \ln (81) = x (- \ln (81)) - 2 \ln (3)$

Simplify $- 2 \ln (3)$ by moving $2$ inside the logarithm.

$- x \ln (3) + \ln (81) = x (- \ln (81)) - \ln (3^{2})$

Raise $3$ to the power of $2$ .

$- x \ln (3) + \ln (81) = x (- \ln (81)) - \ln (9)$

Reorder factors in $x (- \ln (81)) - \ln (9)$ .

$- x \ln (3) + \ln (81) = - x \ln (81) - \ln (9)$

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

$- x \ln (3) + \ln (81) + x \ln (81) + \ln (9) = 0$

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

$- x \ln (3) + x \ln (81) + \ln (81 \cdot 9) = 0$

Multiply $81$ by $9$ .

$- x \ln (3) + x \ln (81) + \ln (729) = 0$

Subtract $\ln (729)$ from both sides of the equation.

$- x \ln (3) + x \ln (81) = - \ln (729)$

Factor $x$ out of $- x \ln (3) + x \ln (81)$ .

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

$x (- 1 \ln (3)) + x \ln (81) = - \ln (729)$

Factor $x$ out of $x \ln (81)$ .

$x (- 1 \ln (3)) + x (\ln (81)) = - \ln (729)$

Factor $x$ out of $x (- 1 \ln (3)) + x (\ln (81))$ .

$x (- 1 \ln (3) + \ln (81)) = - \ln (729)$

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

$x (- \ln (3) + \ln (81)) = - \ln (729)$

Divide each term in $x (- \ln (3) + \ln (81)) = - \ln (729)$ by $- \ln (3) + \ln (81)$ and simplify.

Divide each term in $x (- \ln (3) + \ln (81)) = - \ln (729)$ by $- \ln (3) + \ln (81)$ .

$\frac{x (- \ln (3) + \ln (81))}{- \ln (3) + \ln (81)} = \frac{- \ln (729)}{- \ln (3) + \ln (81)}$

Simplify the left side.

Cancel the common factor of $- \ln (3) + \ln (81)$ .

Cancel the common factor.

$\frac{x (- \ln (3) + \ln (81))}{- \ln (3) + \ln (81)} = \frac{- \ln (729)}{- \ln (3) + \ln (81)}$

Divide $x$ by $1$ .

$x = \frac{- \ln (729)}{- \ln (3) + \ln (81)}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{\ln (729)}{- \ln (3) + \ln (81)}$

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

$x = - \frac{\ln (729)}{- \ln (3) + \ln (81)}$

Factor $- 1$ out of $\ln (81)$ .

$x = - \frac{\ln (729)}{- \ln (3) - 1 (- \ln (81))}$

Factor $- 1$ out of $- \ln (3) - 1 (- \ln (81))$ .

$x = - \frac{\ln (729)}{- (\ln (3) - \ln (81))}$

Simplify the expression.

Rewrite $- (\ln (3) - \ln (81))$ as $- 1 (\ln (3) - \ln (81))$ .

$x = - \frac{\ln (729)}{- 1 (\ln (3) - \ln (81))}$

Move the negative in front of the fraction.

$x = - - \frac{\ln (729)}{\ln (3) - \ln (81)}$

Multiply $- 1$ by $- 1$ .

$x = 1 \frac{\ln (729)}{\ln (3) - \ln (81)}$

Multiply $\frac{\ln (729)}{\ln (3) - \ln (81)}$ by $1$ .

$x = \frac{\ln (729)}{\ln (3) - \ln (81)}$

Step 5

The solution consists of all of the true intervals.

$x < \frac{\ln (729)}{\ln (3) - \ln (81)}$

Step 6

The result can be shown in multiple forms.

Inequality Form:

$x < \frac{\ln (729)}{\ln (3) - \ln (81)}$

Interval Notation:

$(- \infty, \frac{\ln (729)}{\ln (3) - \ln (81)})$

Step 7

Solve the Inequality for x (1/3)^(2x+1)> square root of 27/(3^(x-1))

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