Basic Math Examples

$({(\frac{3}{5})}^{z} {(\frac{5}{3})}^{2 z}) = \frac{125}{27}$

Step 1

Simplify ${(\frac{3}{5})}^{z} {(\frac{5}{3})}^{2 z}$ .

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Step 1.1

Apply basic rules of exponents.

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Step 1.1.1

Apply the product rule to $\frac{3}{5}$ .

$\frac{3^{z}}{5^{z}} {(\frac{5}{3})}^{2 z} = \frac{125}{27}$

Step 1.1.2

Apply the product rule to $\frac{5}{3}$ .

$\frac{3^{z}}{5^{z}} \cdot \frac{5^{2 z}}{3^{2 z}} = \frac{125}{27}$

Step 1.2

Combine.

$\frac{3^{z} \cdot 5^{2 z}}{5^{z} \cdot 3^{2 z}} = \frac{125}{27}$

Step 1.3

Cancel the common factor of $3^{z}$ and $3^{2 z}$ .

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Step 1.3.1

Factor $3^{2 z}$ out of $3^{z} \cdot 5^{2 z}$ .

$\frac{3^{2 z} (3^{- z} \cdot 5^{2 z})}{5^{z} \cdot 3^{2 z}} = \frac{125}{27}$

Step 1.3.2

Cancel the common factors.

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Step 1.3.2.1

Factor $3^{2 z}$ out of $5^{z} \cdot 3^{2 z}$ .

$\frac{3^{2 z} (3^{- z} \cdot 5^{2 z})}{3^{2 z} \cdot 5^{z}} = \frac{125}{27}$

Step 1.3.2.2

Cancel the common factor.

$\frac{3^{2 z} (3^{- z} \cdot 5^{2 z})}{3^{2 z} \cdot 5^{z}} = \frac{125}{27}$

Step 1.3.2.3

Rewrite the expression.

$\frac{3^{- z} \cdot 5^{2 z}}{5^{z}} = \frac{125}{27}$

Step 1.4

Cancel the common factor of $5^{2 z}$ and $5^{z}$ .

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Step 1.4.1

Factor $5^{z}$ out of $3^{- z} \cdot 5^{2 z}$ .

$\frac{5^{z} (3^{- z} \cdot 5^{z})}{5^{z}} = \frac{125}{27}$

Step 1.4.2

Cancel the common factors.

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Step 1.4.2.1

Multiply by $1$ .

$\frac{5^{z} (3^{- z} \cdot 5^{z})}{5^{z} \cdot 1} = \frac{125}{27}$

Step 1.4.2.2

Cancel the common factor.

$\frac{5^{z} (3^{- z} \cdot 5^{z})}{5^{z} \cdot 1} = \frac{125}{27}$

Step 1.4.2.3

Rewrite the expression.

$\frac{3^{- z} \cdot 5^{z}}{1} = \frac{125}{27}$

Step 1.4.2.4

Divide $3^{- z} \cdot 5^{z}$ by $1$ .

$3^{- z} \cdot 5^{z} = \frac{125}{27}$

Step 2

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

$\ln (3^{- z} \cdot 5^{z}) = \ln (\frac{125}{27})$

Step 3

Expand the left side.

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Step 3.1

Rewrite $\ln (3^{- z} \cdot 5^{z})$ as $\ln (3^{- z}) + \ln (5^{z})$ .

$\ln (3^{- z}) + \ln (5^{z}) = \ln (\frac{125}{27})$

Step 3.2

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

$- z \ln (3) + \ln (5^{z}) = \ln (\frac{125}{27})$

Step 3.3

Expand $\ln (5^{z})$ by moving $z$ outside the logarithm.

$- z \ln (3) + z \ln (5) = \ln (\frac{125}{27})$

Step 4

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

$- z \ln (3) + z \ln (5) - \ln (\frac{125}{27}) = 0$

Step 5

Add $\ln (\frac{125}{27})$ to both sides of the equation.

$- z \ln (3) + z \ln (5) = \ln (\frac{125}{27})$

Step 6

Factor $z$ out of $- z \ln (3) + z \ln (5)$ .

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Step 6.1

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

$z (- 1 \ln (3)) + z \ln (5) = \ln (\frac{125}{27})$

Step 6.2

Factor $z$ out of $z \ln (5)$ .

$z (- 1 \ln (3)) + z (\ln (5)) = \ln (\frac{125}{27})$

Step 6.3

Factor $z$ out of $z (- 1 \ln (3)) + z (\ln (5))$ .

$z (- 1 \ln (3) + \ln (5)) = \ln (\frac{125}{27})$

Step 7

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

$z (- \ln (3) + \ln (5)) = \ln (\frac{125}{27})$

Step 8

Divide each term in $z (- \ln (3) + \ln (5)) = \ln (\frac{125}{27})$ by $- \ln (3) + \ln (5)$ and simplify.

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Step 8.1

Divide each term in $z (- \ln (3) + \ln (5)) = \ln (\frac{125}{27})$ by $- \ln (3) + \ln (5)$ .

$\frac{z (- \ln (3) + \ln (5))}{- \ln (3) + \ln (5)} = \frac{\ln (\frac{125}{27})}{- \ln (3) + \ln (5)}$

Step 8.2

Simplify the left side.

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Step 8.2.1

Cancel the common factor of $- \ln (3) + \ln (5)$ .

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Step 8.2.1.1

Cancel the common factor.

$\frac{z (- \ln (3) + \ln (5))}{- \ln (3) + \ln (5)} = \frac{\ln (\frac{125}{27})}{- \ln (3) + \ln (5)}$

Step 8.2.1.2

Divide $z$ by $1$ .

$z = \frac{\ln (\frac{125}{27})}{- \ln (3) + \ln (5)}$

Step 8.3

Simplify the right side.

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Step 8.3.1

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

$z = \frac{\ln (\frac{125}{27})}{- \ln (3) + \ln (5)}$

Step 8.3.2

Factor $- 1$ out of $\ln (5)$ .

$z = \frac{\ln (\frac{125}{27})}{- \ln (3) - 1 (- \ln (5))}$

Step 8.3.3

Factor $- 1$ out of $- \ln (3) - 1 (- \ln (5))$ .

$z = \frac{\ln (\frac{125}{27})}{- (\ln (3) - \ln (5))}$

Step 8.3.4

Rewrite negatives.

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Step 8.3.4.1

Rewrite $- (\ln (3) - \ln (5))$ as $- 1 (\ln (3) - \ln (5))$ .

$z = \frac{\ln (\frac{125}{27})}{- 1 (\ln (3) - \ln (5))}$

Step 8.3.4.2

Move the negative in front of the fraction.

$z = - \frac{\ln (\frac{125}{27})}{\ln (3) - \ln (5)}$

Step 9

The result can be shown in multiple forms.

Exact Form:

$z = - \frac{\ln (\frac{125}{27})}{\ln (3) - \ln (5)}$

Decimal Form:

$z = 3$