Basic Math Examples

\frac{\sqrt{3^{y} + 3^{y} + 3^{y}}}{\sqrt{6^{y} + 6^{y} + 6^{y}}} = \frac{1}{64}

$\frac{\sqrt{3^{y} + 3^{y} + 3^{y}}}{\sqrt{6^{y} + 6^{y} + 6^{y}}} = \frac{1}{64}$

Step 1

Cross multiply.

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Step 1.1

Cross multiply by setting the product of the numerator of the right side and the denominator of the left side equal to the product of the numerator of the left side and the denominator of the right side.

1 \cdot (\sqrt{6^{y} + 6^{y} + 6^{y}}) = \sqrt{3^{y} + 3^{y} + 3^{y}} \cdot (64)

$1 \cdot (\sqrt{6^{y} + 6^{y} + 6^{y}}) = \sqrt{3^{y} + 3^{y} + 3^{y}} \cdot (64)$

Step 1.2

Simplify the left side.

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Step 1.2.1

Simplify

1 \cdot (\sqrt{6^{y} + 6^{y} + 6^{y}})

$1 \cdot (\sqrt{6^{y} + 6^{y} + 6^{y}})$ .

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Step 1.2.1.1

Multiply

\sqrt{6^{y} + 6^{y} + 6^{y}}

$\sqrt{6^{y} + 6^{y} + 6^{y}}$ by

1

$1$ .

\sqrt{6^{y} + 6^{y} + 6^{y}} = \sqrt{3^{y} + 3^{y} + 3^{y}} \cdot (64)

$\sqrt{6^{y} + 6^{y} + 6^{y}} = \sqrt{3^{y} + 3^{y} + 3^{y}} \cdot (64)$

Step 1.2.1.2

Add

6^{y}

$6^{y}$ and

6^{y}

$6^{y}$ .

\sqrt{2 \cdot 6^{y} + 6^{y}} = \sqrt{3^{y} + 3^{y} + 3^{y}} \cdot (64)

$\sqrt{2 \cdot 6^{y} + 6^{y}} = \sqrt{3^{y} + 3^{y} + 3^{y}} \cdot (64)$

Step 1.2.1.3

Add

2 \cdot 6^{y}

$2 \cdot 6^{y}$ and

6^{y}

$6^{y}$ .

\sqrt{3 \cdot 6^{y}} = \sqrt{3^{y} + 3^{y} + 3^{y}} \cdot (64)

$\sqrt{3 \cdot 6^{y}} = \sqrt{3^{y} + 3^{y} + 3^{y}} \cdot (64)$

\sqrt{3 \cdot 6^{y}} = \sqrt{3^{y} + 3^{y} + 3^{y}} \cdot (64)

$\sqrt{3 \cdot 6^{y}} = \sqrt{3^{y} + 3^{y} + 3^{y}} \cdot (64)$

\sqrt{3 \cdot 6^{y}} = \sqrt{3^{y} + 3^{y} + 3^{y}} \cdot (64)

$\sqrt{3 \cdot 6^{y}} = \sqrt{3^{y} + 3^{y} + 3^{y}} \cdot (64)$

Step 1.3

Simplify the right side.

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Step 1.3.1

Simplify

\sqrt{3^{y} + 3^{y} + 3^{y}} \cdot (64)

$\sqrt{3^{y} + 3^{y} + 3^{y}} \cdot (64)$ .

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Step 1.3.1.1

Add

3^{y}

$3^{y}$ and

3^{y}

$3^{y}$ .

\sqrt{3 \cdot 6^{y}} = \sqrt{2 \cdot 3^{y} + 3^{y}} \cdot 64

$\sqrt{3 \cdot 6^{y}} = \sqrt{2 \cdot 3^{y} + 3^{y}} \cdot 64$

Step 1.3.1.2

Add

2 \cdot 3^{y}

$2 \cdot 3^{y}$ and

3^{y}

$3^{y}$ .

\sqrt{3 \cdot 6^{y}} = \sqrt{3 \cdot 3^{y}} \cdot 64

$\sqrt{3 \cdot 6^{y}} = \sqrt{3 \cdot 3^{y}} \cdot 64$

Step 1.3.1.3

Multiply

3

$3$ by

3^{y}

$3^{y}$ .

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Step 1.3.1.3.1

Raise

3

$3$ to the power of

1

$1$ .

\sqrt{3 \cdot 6^{y}} = \sqrt{3^{1} \cdot 3^{y}} \cdot 64

$\sqrt{3 \cdot 6^{y}} = \sqrt{3^{1} \cdot 3^{y}} \cdot 64$

Step 1.3.1.3.2

Use the power rule

a^{m} a^{n} = a^{m + n}

$a^{m} a^{n} = a^{m + n}$ to combine exponents.

\sqrt{3 \cdot 6^{y}} = \sqrt{3^{1 + y}} \cdot 64

$\sqrt{3 \cdot 6^{y}} = \sqrt{3^{1 + y}} \cdot 64$

\sqrt{3 \cdot 6^{y}} = \sqrt{3^{1 + y}} \cdot 64

$\sqrt{3 \cdot 6^{y}} = \sqrt{3^{1 + y}} \cdot 64$

Step 1.3.1.4

Move

64

$64$ to the left of

\sqrt{3^{1 + y}}

$\sqrt{3^{1 + y}}$ .

\sqrt{3 \cdot 6^{y}} = 64 \sqrt{3^{1 + y}}

$\sqrt{3 \cdot 6^{y}} = 64 \sqrt{3^{1 + y}}$

\sqrt{3 \cdot 6^{y}} = 64 \sqrt{3^{1 + y}}

$\sqrt{3 \cdot 6^{y}} = 64 \sqrt{3^{1 + y}}$

\sqrt{3 \cdot 6^{y}} = 64 \sqrt{3^{1 + y}}

$\sqrt{3 \cdot 6^{y}} = 64 \sqrt{3^{1 + y}}$

\sqrt{3 \cdot 6^{y}} = 64 \sqrt{3^{1 + y}}

$\sqrt{3 \cdot 6^{y}} = 64 \sqrt{3^{1 + y}}$

Step 2

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

{\sqrt{3 \cdot 6^{y}}}^{2} = {(64 \sqrt{3^{1 + y}})}^{2}

${\sqrt{3 \cdot 6^{y}}}^{2} = {(64 \sqrt{3^{1 + y}})}^{2}$

Step 3

Simplify each side of the equation.

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Step 3.1

Use

\sqrt[n]{a^{x}} = a^{\frac{x}{n}}

$\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite

\sqrt{3 \cdot 6^{y}}

$\sqrt{3 \cdot 6^{y}}$ as

{(3 \cdot 6^{y})}^{\frac{1}{2}}

${(3 \cdot 6^{y})}^{\frac{1}{2}}$ .

{({(3 \cdot 6^{y})}^{\frac{1}{2}})}^{2} = {(64 \sqrt{3^{1 + y}})}^{2}

${({(3 \cdot 6^{y})}^{\frac{1}{2}})}^{2} = {(64 \sqrt{3^{1 + y}})}^{2}$

Step 3.2

Simplify the left side.

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Step 3.2.1

Simplify

{({(3 \cdot 6^{y})}^{\frac{1}{2}})}^{2}

${({(3 \cdot 6^{y})}^{\frac{1}{2}})}^{2}$ .

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Step 3.2.1.1

Multiply the exponents in

{({(3 \cdot 6^{y})}^{\frac{1}{2}})}^{2}

${({(3 \cdot 6^{y})}^{\frac{1}{2}})}^{2}$ .

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Step 3.2.1.1.1

Apply the power rule and multiply exponents,

{(a^{m})}^{n} = a^{m n}

${(a^{m})}^{n} = a^{m n}$ .

{(3 \cdot 6^{y})}^{\frac{1}{2} \cdot 2} = {(64 \sqrt{3^{1 + y}})}^{2}

${(3 \cdot 6^{y})}^{\frac{1}{2} \cdot 2} = {(64 \sqrt{3^{1 + y}})}^{2}$

Step 3.2.1.1.2

Cancel the common factor of

2

$2$ .

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Step 3.2.1.1.2.1

Cancel the common factor.

${(3 \cdot 6^{y})}^{\frac{1}{2} \cdot 2} = {(64 \sqrt{3^{1 + y}})}^{2}$

Step 3.2.1.1.2.2

Rewrite the expression.

${(3 \cdot 6^{y})}^{1} = {(64 \sqrt{3^{1 + y}})}^{2}$

Step 3.2.1.2

Simplify.

$3 \cdot 6^{y} = {(64 \sqrt{3^{1 + y}})}^{2}$

Step 3.3

Simplify the right side.

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Step 3.3.1

Simplify

${(64 \sqrt{3^{1 + y}})}^{2}$ .

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Step 3.3.1.1

Simplify the expression.

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Step 3.3.1.1.1

Apply the product rule to

$64 \sqrt{3^{1 + y}}$ .

$3 \cdot 6^{y} = 64^{2} {\sqrt{3^{1 + y}}}^{2}$

Step 3.3.1.1.2

Raise

$64$ to the power of

$2$ .

$3 \cdot 6^{y} = 4096 {\sqrt{3^{1 + y}}}^{2}$

Step 3.3.1.2

Rewrite

${\sqrt{3^{1 + y}}}^{2}$ as

$3^{1 + y}$ .

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Step 3.3.1.2.1

Use

$\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite

$\sqrt{3^{1 + y}}$ as

$3^{\frac{1 + y}{2}}$ .

$3 \cdot 6^{y} = 4096 {(3^{\frac{1 + y}{2}})}^{2}$

Step 3.3.1.2.2

Apply the power rule and multiply exponents,

${(a^{m})}^{n} = a^{m n}$ .

$3 \cdot 6^{y} = 4096 \cdot 3^{\frac{1 + y}{2} \cdot 2}$

Step 3.3.1.2.3

Combine

$\frac{1 + y}{2}$ and

$2$ .

$3 \cdot 6^{y} = 4096 \cdot 3^{\frac{(1 + y) \cdot 2}{2}}$

Step 3.3.1.2.4

Cancel the common factor of

$2$ .

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Step 3.3.1.2.4.1

Cancel the common factor.

$3 \cdot 6^{y} = 4096 \cdot 3^{\frac{(1 + y) \cdot 2}{2}}$

Step 3.3.1.2.4.2

Divide

$1 + y$ by

$1$ .

$3 \cdot 6^{y} = 4096 \cdot 3^{1 + y}$

Step 4

Solve for

$y$ .

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Step 4.1

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

$\ln (3 \cdot 6^{y}) = \ln (4096 \cdot 3^{1 + y})$

Step 4.2

Expand the left side.

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Step 4.2.1

Rewrite

$\ln (3 \cdot 6^{y})$ as

$\ln (3) + \ln (6^{y})$ .

$\ln (3) + \ln (6^{y}) = \ln (4096 \cdot 3^{1 + y})$

Step 4.2.2

Expand

$\ln (6^{y})$ by moving

$y$ outside the logarithm.

$\ln (3) + y \ln (6) = \ln (4096 \cdot 3^{1 + y})$

Step 4.3

Expand the right side.

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Step 4.3.1

Rewrite

$\ln (4096 \cdot 3^{1 + y})$ as

$\ln (4096) + \ln (3^{1 + y})$ .

$\ln (3) + y \ln (6) = \ln (4096) + \ln (3^{1 + y})$

Step 4.3.2

Expand

$\ln (3^{1 + y})$ by moving

$1 + y$ outside the logarithm.

$\ln (3) + y \ln (6) = \ln (4096) + (1 + y) \ln (3)$

Step 4.4

Simplify the right side.

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Step 4.4.1

Simplify

$\ln (4096) + (1 + y) \ln (3)$ .

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Step 4.4.1.1

Simplify each term.

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Step 4.4.1.1.1

Apply the distributive property.

$\ln (3) + y \ln (6) = \ln (4096) + 1 \ln (3) + y \ln (3)$

Step 4.4.1.1.2

Multiply

$\ln (3)$ by

$1$ .

$\ln (3) + y \ln (6) = \ln (4096) + \ln (3) + y \ln (3)$

Step 4.4.1.2

Use the product property of logarithms,

$\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\ln (3) + y \ln (6) = \ln (4096 \cdot 3) + y \ln (3)$

Step 4.4.1.3

Multiply

$4096$ by

$3$ .

$\ln (3) + y \ln (6) = \ln (12288) + y \ln (3)$

Step 4.5

Reorder

$\ln (3)$ and

$y \ln (6)$ .

$y \ln (6) + \ln (3) = \ln (12288) + y \ln (3)$

Step 4.6

Reorder

$\ln (12288)$ and

$y \ln (3)$ .

$y \ln (6) + \ln (3) = y \ln (3) + \ln (12288)$

Step 4.7

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

$y \ln (6) + \ln (3) - y \ln (3) - \ln (12288) = 0$

Step 4.8

Use the quotient property of logarithms,

$\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$y \ln (6) - y \ln (3) + \ln (\frac{3}{12288}) = 0$

Step 4.9

Cancel the common factor of

$3$ and

$12288$ .

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Step 4.9.1

Factor

$3$ out of

$3$ .

$y \ln (6) - y \ln (3) + \ln (\frac{3 (1)}{12288}) = 0$

Step 4.9.2

Cancel the common factors.

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Step 4.9.2.1

Factor

$3$ out of

$12288$ .

$y \ln (6) - y \ln (3) + \ln (\frac{3 \cdot 1}{3 \cdot 4096}) = 0$

Step 4.9.2.2

Cancel the common factor.

$y \ln (6) - y \ln (3) + \ln (\frac{3 \cdot 1}{3 \cdot 4096}) = 0$

Step 4.9.2.3

Rewrite the expression.

$y \ln (6) - y \ln (3) + \ln (\frac{1}{4096}) = 0$

Step 4.10

Subtract

$\ln (\frac{1}{4096})$ from both sides of the equation.

$y \ln (6) - y \ln (3) = - \ln (\frac{1}{4096})$

Step 4.11

Factor

$y$ out of

$y \ln (6) - y \ln (3)$ .

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Step 4.11.1

Factor

$y$ out of

$y \ln (6)$ .

$y (\ln (6)) - y \ln (3) = - \ln (\frac{1}{4096})$

Step 4.11.2

Factor

$y$ out of

$- y \ln (3)$ .

$y (\ln (6)) + y (- 1 \ln (3)) = - \ln (\frac{1}{4096})$

Step 4.11.3

Factor

$y$ out of

$y (\ln (6)) + y (- 1 \ln (3))$ .

$y (\ln (6) - 1 \ln (3)) = - \ln (\frac{1}{4096})$

Step 4.12

Rewrite

$- 1 \ln (3)$ as

$- \ln (3)$ .

$y (\ln (6) - \ln (3)) = - \ln (\frac{1}{4096})$

Step 4.13

Divide each term in

$y (\ln (6) - \ln (3)) = - \ln (\frac{1}{4096})$ by

$\ln (6) - \ln (3)$ and simplify.

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Step 4.13.1

Divide each term in

$y (\ln (6) - \ln (3)) = - \ln (\frac{1}{4096})$ by

$\ln (6) - \ln (3)$ .

$\frac{y (\ln (6) - \ln (3))}{\ln (6) - \ln (3)} = \frac{- \ln (\frac{1}{4096})}{\ln (6) - \ln (3)}$

Step 4.13.2

Simplify the left side.

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Step 4.13.2.1

Cancel the common factor of

$\ln (6) - \ln (3)$ .

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Step 4.13.2.1.1

Cancel the common factor.

$\frac{y (\ln (6) - \ln (3))}{\ln (6) - \ln (3)} = \frac{- \ln (\frac{1}{4096})}{\ln (6) - \ln (3)}$

Step 4.13.2.1.2

Divide

$y$ by

$1$ .

$y = \frac{- \ln (\frac{1}{4096})}{\ln (6) - \ln (3)}$

Step 4.13.3

Simplify the right side.

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Step 4.13.3.1

Move the negative in front of the fraction.

$y = - \frac{\ln (\frac{1}{4096})}{\ln (6) - \ln (3)}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$y = - \frac{\ln (\frac{1}{4096})}{\ln (6) - \ln (3)}$

Decimal Form:

$y = 12$