Solve for x (49/81)^(x+1)>=9/7

${(\frac{49}{81})}^{x + 1} \geq \frac{9}{7}$

Step 1

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln ({(\frac{49}{81})}^{x + 1}) \geq \ln (\frac{9}{7})$

Step 2

Expand the left side.

Expand $\ln ({(\frac{49}{81})}^{x + 1})$ by moving $x + 1$ outside the logarithm.

$(x + 1) \ln (\frac{49}{81}) \geq \ln (\frac{9}{7})$

Rewrite $\ln (\frac{49}{81})$ as $\ln (49) - \ln (81)$ .

$(x + 1) (\ln (49) - \ln (81)) \geq \ln (\frac{9}{7})$

Rewrite $\ln (81)$ as $\ln (3^{4})$ .

$(x + 1) (\ln (49) - \ln (3^{4})) \geq \ln (\frac{9}{7})$

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

$(x + 1) (\ln (49) - (4 \ln (3))) \geq \ln (\frac{9}{7})$

Multiply $4$ by $- 1$ .

$(x + 1) (\ln (49) - 4 \ln (3)) \geq \ln (\frac{9}{7})$

Step 3

Simplify the left side.

Simplify $(x + 1) (\ln (49) - 4 \ln (3))$ .

Expand $(x + 1) (\ln (49) - 4 \ln (3))$ using the FOIL Method.

Apply the distributive property.

$x (\ln (49) - 4 \ln (3)) + 1 (\ln (49) - 4 \ln (3)) \geq \ln (\frac{9}{7})$

Apply the distributive property.

$x \ln (49) + x (- 4 \ln (3)) + 1 (\ln (49) - 4 \ln (3)) \geq \ln (\frac{9}{7})$

Apply the distributive property.

$x \ln (49) + x (- 4 \ln (3)) + 1 \ln (49) + 1 (- 4 \ln (3)) \geq \ln (\frac{9}{7})$

Simplify terms.

Simplify each term.

Multiply $\ln (49)$ by $1$ .

$x \ln (49) + x (- 4 \ln (3)) + \ln (49) + 1 (- 4 \ln (3)) \geq \ln (\frac{9}{7})$

Multiply $- 4 \ln (3)$ by $1$ .

$x \ln (49) + x (- 4 \ln (3)) + \ln (49) - 4 \ln (3) \geq \ln (\frac{9}{7})$

Reorder factors in $x \ln (49) + x (- 4 \ln (3)) + \ln (49) - 4 \ln (3)$ .

$x \ln (49) - 4 x \ln (3) + \ln (49) - 4 \ln (3) \geq \ln (\frac{9}{7})$

Step 4

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

$x \ln (49) - 4 x \ln (3) + \ln (49) - 4 \ln (3) - \ln (\frac{9}{7}) = 0$

Step 5

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

$x \ln (49) - 4 x \ln (3) + \ln (\frac{49}{\frac{9}{7}}) - 4 \ln (3) = 0$

Step 6

Simplify each term.

Multiply the numerator by the reciprocal of the denominator.

$x \ln (49) - 4 x \ln (3) + \ln (49 (\frac{7}{9})) - 4 \ln (3) = 0$

Multiply $49 (\frac{7}{9})$ .

Combine $49$ and $\frac{7}{9}$ .

$x \ln (49) - 4 x \ln (3) + \ln (\frac{49 \cdot 7}{9}) - 4 \ln (3) = 0$

Multiply $49$ by $7$ .

$x \ln (49) - 4 x \ln (3) + \ln (\frac{343}{9}) - 4 \ln (3) = 0$

Step 7

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

Subtract $\ln (\frac{343}{9})$ from both sides of the equation.

$x \ln (49) - 4 x \ln (3) - 4 \ln (3) = - \ln (\frac{343}{9})$

Add $4 \ln (3)$ to both sides of the equation.

$x \ln (49) - 4 x \ln (3) = - \ln (\frac{343}{9}) + 4 \ln (3)$

Step 8

Factor $x$ out of $x \ln (49) - 4 x \ln (3)$ .

Factor $x$ out of $x \ln (49)$ .

$x (\ln (49)) - 4 x \ln (3) = - \ln (\frac{343}{9}) + 4 \ln (3)$

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

$x (\ln (49)) + x (- 4 \ln (3)) = - \ln (\frac{343}{9}) + 4 \ln (3)$

Factor $x$ out of $x (\ln (49)) + x (- 4 \ln (3))$ .

$x (\ln (49) - 4 \ln (3)) = - \ln (\frac{343}{9}) + 4 \ln (3)$

Step 9

Divide each term in $x (\ln (49) - 4 \ln (3)) = - \ln (\frac{343}{9}) + 4 \ln (3)$ by $\ln (49) - 4 \ln (3)$ and simplify.

Divide each term in $x (\ln (49) - 4 \ln (3)) = - \ln (\frac{343}{9}) + 4 \ln (3)$ by $\ln (49) - 4 \ln (3)$ .

$\frac{x (\ln (49) - 4 \ln (3))}{\ln (49) - 4 \ln (3)} = \frac{- \ln (\frac{343}{9})}{\ln (49) - 4 \ln (3)} + \frac{4 \ln (3)}{\ln (49) - 4 \ln (3)}$

Simplify the left side.

Cancel the common factor of $\ln (49) - 4 \ln (3)$ .

Cancel the common factor.

$\frac{x (\ln (49) - 4 \ln (3))}{\ln (49) - 4 \ln (3)} = \frac{- \ln (\frac{343}{9})}{\ln (49) - 4 \ln (3)} + \frac{4 \ln (3)}{\ln (49) - 4 \ln (3)}$

Divide $x$ by $1$ .

$x = \frac{- \ln (\frac{343}{9})}{\ln (49) - 4 \ln (3)} + \frac{4 \ln (3)}{\ln (49) - 4 \ln (3)}$

Simplify the right side.

Combine the numerators over the common denominator.

$x = \frac{- \ln (\frac{343}{9}) + 4 \ln (3)}{\ln (49) - 4 \ln (3)}$

Factor $- 1$ out of $- \ln (\frac{343}{9})$ .

$x = \frac{- \ln (\frac{343}{9}) + 4 \ln (3)}{\ln (49) - 4 \ln (3)}$

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

$x = \frac{- \ln (\frac{343}{9}) - (- 4 \ln (3))}{\ln (49) - 4 \ln (3)}$

Factor $- 1$ out of $- \ln (\frac{343}{9}) - (- 4 \ln (3))$ .

$x = \frac{- (\ln (\frac{343}{9}) - 4 \ln (3))}{\ln (49) - 4 \ln (3)}$

Simplify the expression.

Rewrite $- (\ln (\frac{343}{9}) - 4 \ln (3))$ as $- 1 (\ln (\frac{343}{9}) - 4 \ln (3))$ .

$x = \frac{- 1 (\ln (\frac{343}{9}) - 4 \ln (3))}{\ln (49) - 4 \ln (3)}$

Move the negative in front of the fraction.

$x = - \frac{\ln (\frac{343}{9}) - 4 \ln (3)}{\ln (49) - 4 \ln (3)}$

Step 10

The solution consists of all of the true intervals.

$x \leq - \frac{\ln (\frac{343}{9}) - 4 \ln (3)}{\ln (49) - 4 \ln (3)}$

Step 11

The result can be shown in multiple forms.

Inequality Form:

$x \leq - \frac{\ln (\frac{343}{9}) - 4 \ln (3)}{\ln (49) - 4 \ln (3)}$

Interval Notation:

$(- \infty, - \frac{\ln (\frac{343}{9}) - 4 \ln (3)}{\ln (49) - 4 \ln (3)}]$

Step 12

Solve for x (49/81)^(x+1)>=9/7

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