Trigonometry Examples

$\log ({(81)}^{y}) < \log ({(27)}^{y + 3})$

Step 1

Convert the inequality to an equality.

$\log (81^{y}) = \log (27^{y + 3})$

Step 2

Solve the equation.

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Step 2.1

Take the log of both sides of the equation.

$\ln (\log (81^{y})) = \ln (\log (27^{y + 3}))$

Step 2.2

Expand $\log (81^{y})$ by moving $y$ outside the logarithm.

$\ln (y \log (81)) = \ln (\log (27^{y + 3}))$

Step 2.3

Rewrite $\ln (y \log (81))$ as $\ln (y) + \ln (\log (81))$ .

$\ln (y) + \ln (\log (81)) = \ln (\log (27^{y + 3}))$

Step 2.4

Expand $\log (27^{y + 3})$ by moving $y + 3$ outside the logarithm.

$\ln (y) + \ln (\log (81)) = \ln ((y + 3) \log (27))$

Step 2.5

Rewrite $\ln ((y + 3) \log (27))$ as $\ln (y + 3) + \ln (\log (27))$ .

$\ln (y) + \ln (\log (81)) = \ln (y + 3) + \ln (\log (27))$

Step 2.6

Solve the equation for $y$ .

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Step 2.6.1

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

$\ln (y \log (81)) = \ln (y + 3) + \ln (\log (27))$

Step 2.6.2

Simplify the right side.

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Step 2.6.2.1

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

$\ln (y \log (81)) = \ln ((y + 3) \log (27))$

Step 2.6.2.2

Apply the distributive property.

$\ln (y \log (81)) = \ln (y \log (27) + 3 \log (27))$

Step 2.6.3

Simplify $3 \log (27)$ by moving $3$ inside the logarithm.

$\ln (y \log (81)) = \ln (y \log (27) + \log (27^{3}))$

Step 2.6.4

For the equation to be equal, the argument of the logarithms on both sides of the equation must be equal.

$y \log (81) = y \log (27) + \log (27^{3})$

Step 2.6.5

Solve for $y$ .

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Step 2.6.5.1

Simplify the right side.

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Step 2.6.5.1.1

Raise $27$ to the power of $3$ .

$y \log (81) = y \log (27) + \log (19683)$

Step 2.6.5.2

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

$y \log (81) - y \log (27) - \log (19683) = 0$

Step 2.6.5.3

Add $\log (19683)$ to both sides of the equation.

$y \log (81) - y \log (27) = \log (19683)$

Step 2.6.5.4

Factor $y$ out of $y \log (81) - y \log (27)$ .

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Step 2.6.5.4.1

Factor $y$ out of $y \log (81)$ .

$y (\log (81)) - y \log (27) = \log (19683)$

Step 2.6.5.4.2

Factor $y$ out of $- y \log (27)$ .

$y (\log (81)) + y (- 1 \log (27)) = \log (19683)$

Step 2.6.5.4.3

Factor $y$ out of $y (\log (81)) + y (- 1 \log (27))$ .

$y (\log (81) - 1 \log (27)) = \log (19683)$

Step 2.6.5.5

Rewrite $- 1 \log (27)$ as $- \log (27)$ .

$y (\log (81) - \log (27)) = \log (19683)$

Step 2.6.5.6

Divide each term in $y (\log (81) - \log (27)) = \log (19683)$ by $\log (81) - \log (27)$ and simplify.

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Step 2.6.5.6.1

Divide each term in $y (\log (81) - \log (27)) = \log (19683)$ by $\log (81) - \log (27)$ .

$\frac{y (\log (81) - \log (27))}{\log (81) - \log (27)} = \frac{\log (19683)}{\log (81) - \log (27)}$

Step 2.6.5.6.2

Simplify the left side.

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Step 2.6.5.6.2.1

Cancel the common factor of $\log (81) - \log (27)$ .

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Step 2.6.5.6.2.1.1

Cancel the common factor.

$\frac{y (\log (81) - \log (27))}{\log (81) - \log (27)} = \frac{\log (19683)}{\log (81) - \log (27)}$

Step 2.6.5.6.2.1.2

Divide $y$ by $1$ .

$y = \frac{\log (19683)}{\log (81) - \log (27)}$

Step 3

The solution consists of all of the true intervals.

$y < \frac{\log (19683)}{\log (81) - \log (27)}$

Step 4

The result can be shown in multiple forms.

Inequality Form:

$y < \frac{\log (19683)}{\log (81) - \log (27)}$

Interval Notation:

$(- \infty, \frac{\log (19683)}{\log (81) - \log (27)})$

Step 5