Solve for d (cy^2)(4y^d)=-20y^3

$(c y^{2}) (4 y^{d}) = - 20 y^{3}$

Step 1

Simplify.

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$c (4 y^{2 + d}) = - 20 y^{3}$

Remove parentheses.

$c \cdot 4 y^{2 + d} = - 20 y^{3}$

Step 2

Divide each term in $(c \cdot (4 y^{2 + d})) = - 20 y^{3}$ by $c y^{2} \cdot 4$ and simplify.

Divide each term in $(c \cdot (4 y^{2 + d})) = - 20 y^{3}$ by $c y^{2} \cdot 4$ .

$\frac{(c \cdot (4 y^{2 + d}))}{c y^{2} \cdot 4} = \frac{- 20 y^{3}}{c y^{2} \cdot 4}$

Simplify the left side.

Cancel the common factor of $c$ .

Cancel the common factor.

$\frac{c \cdot 4 y^{2 + d}}{c y^{2} \cdot 4} = \frac{- 20 y^{3}}{c y^{2} \cdot 4}$

Rewrite the expression.

$\frac{4 y^{2 + d}}{y^{2} \cdot 4} = \frac{- 20 y^{3}}{c y^{2} \cdot 4}$

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 y^{2 + d}}{y^{2} \cdot 4} = \frac{- 20 y^{3}}{c y^{2} \cdot 4}$

Rewrite the expression.

$\frac{y^{2 + d}}{y^{2}} = \frac{- 20 y^{3}}{c y^{2} \cdot 4}$

Cancel the common factor of $y^{2 + d}$ and $y^{2}$ .

Factor $y^{2}$ out of $y^{2 + d}$ .

$\frac{y^{2} y^{d}}{y^{2}} = \frac{- 20 y^{3}}{c y^{2} \cdot 4}$

Cancel the common factors.

Multiply by $1$ .

$\frac{y^{2} y^{d}}{y^{2} \cdot 1} = \frac{- 20 y^{3}}{c y^{2} \cdot 4}$

Cancel the common factor.

$\frac{y^{2} y^{d}}{y^{2} \cdot 1} = \frac{- 20 y^{3}}{c y^{2} \cdot 4}$

Rewrite the expression.

$\frac{y^{d}}{1} = \frac{- 20 y^{3}}{c y^{2} \cdot 4}$

Divide $y^{d}$ by $1$ .

$y^{d} = \frac{- 20 y^{3}}{c y^{2} \cdot 4}$

Simplify the right side.

Cancel the common factor of $- 20$ and $4$ .

Factor $4$ out of $- 20 y^{3}$ .

$y^{d} = \frac{4 (- 5 y^{3})}{c y^{2} \cdot 4}$

Cancel the common factors.

Factor $4$ out of $c y^{2} \cdot 4$ .

$y^{d} = \frac{4 (- 5 y^{3})}{4 \cdot (c y^{2})}$

Cancel the common factor.

$y^{d} = \frac{4 (- 5 y^{3})}{4 \cdot (c y^{2})}$

Rewrite the expression.

$y^{d} = \frac{- 5 y^{3}}{c y^{2}}$

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

Factor $y^{2}$ out of $- 5 y^{3}$ .

$y^{d} = \frac{y^{2} (- 5 y)}{c y^{2}}$

Cancel the common factors.

Factor $y^{2}$ out of $c y^{2}$ .

$y^{d} = \frac{y^{2} (- 5 y)}{y^{2} c}$

Cancel the common factor.

$y^{d} = \frac{y^{2} (- 5 y)}{y^{2} c}$

Rewrite the expression.

$y^{d} = \frac{- 5 y}{c}$

Move the negative in front of the fraction.

$y^{d} = - \frac{5 y}{c}$

Step 3

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

$\ln (y^{d}) = \ln (- \frac{5 y}{c})$

Step 4

Expand $\ln (y^{d})$ by moving $d$ outside the logarithm.

$d \ln (y) = \ln (- \frac{5 y}{c})$

Step 5

Divide each term in $d \ln (y) = \ln (- \frac{5 y}{c})$ by $\ln (y)$ and simplify.

Divide each term in $d \ln (y) = \ln (- \frac{5 y}{c})$ by $\ln (y)$ .

$\frac{d \ln (y)}{\ln (y)} = \frac{\ln (- \frac{5 y}{c})}{\ln (y)}$

Simplify the left side.

Cancel the common factor of $\ln (y)$ .

Cancel the common factor.

$\frac{d \ln (y)}{\ln (y)} = \frac{\ln (- \frac{5 y}{c})}{\ln (y)}$

Divide $d$ by $1$ .

$d = \frac{\ln (- \frac{5 y}{c})}{\ln (y)}$

Solve for d (cy^2)(4y^d)=-20y^3

Related Questions

Support LCMGCF.com by disabling your adblocker.