Solve for r 2700=2200(1+r)^5

$2700 = 2200 {(1 + r)}^{5}$

Step 1

Rewrite the equation as $2200 {(1 + r)}^{5} = 2700$ .

$2200 {(1 + r)}^{5} = 2700$

Step 2

Divide each term in $2200 {(1 + r)}^{5} = 2700$ by $2200$ and simplify.

Divide each term in $2200 {(1 + r)}^{5} = 2700$ by $2200$ .

$\frac{2200 {(1 + r)}^{5}}{2200} = \frac{2700}{2200}$

Simplify the left side.

Cancel the common factor of $2200$ .

Cancel the common factor.

$\frac{2200 {(1 + r)}^{5}}{2200} = \frac{2700}{2200}$

Divide ${(1 + r)}^{5}$ by $1$ .

${(1 + r)}^{5} = \frac{2700}{2200}$

Simplify the right side.

Cancel the common factor of $2700$ and $2200$ .

Factor $100$ out of $2700$ .

${(1 + r)}^{5} = \frac{100 (27)}{2200}$

Cancel the common factors.

Factor $100$ out of $2200$ .

${(1 + r)}^{5} = \frac{100 \cdot 27}{100 \cdot 22}$

Cancel the common factor.

${(1 + r)}^{5} = \frac{100 \cdot 27}{100 \cdot 22}$

Rewrite the expression.

${(1 + r)}^{5} = \frac{27}{22}$

Step 3

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$1 + r = \sqrt[5]{\frac{27}{22}}$

Step 4

Simplify $\sqrt[5]{\frac{27}{22}}$ .

Rewrite $\sqrt[5]{\frac{27}{22}}$ as $\frac{\sqrt[5]{27}}{\sqrt[5]{22}}$ .

$1 + r = \frac{\sqrt[5]{27}}{\sqrt[5]{22}}$

Multiply $\frac{\sqrt[5]{27}}{\sqrt[5]{22}}$ by $\frac{{\sqrt[5]{22}}^{4}}{{\sqrt[5]{22}}^{4}}$ .

$1 + r = \frac{\sqrt[5]{27}}{\sqrt[5]{22}} \cdot \frac{{\sqrt[5]{22}}^{4}}{{\sqrt[5]{22}}^{4}}$

Combine and simplify the denominator.

Multiply $\frac{\sqrt[5]{27}}{\sqrt[5]{22}}$ by $\frac{{\sqrt[5]{22}}^{4}}{{\sqrt[5]{22}}^{4}}$ .

$1 + r = \frac{\sqrt[5]{27} {\sqrt[5]{22}}^{4}}{\sqrt[5]{22} {\sqrt[5]{22}}^{4}}$

Raise $\sqrt[5]{22}$ to the power of $1$ .

$1 + r = \frac{\sqrt[5]{27} {\sqrt[5]{22}}^{4}}{{\sqrt[5]{22}}^{1} {\sqrt[5]{22}}^{4}}$

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

$1 + r = \frac{\sqrt[5]{27} {\sqrt[5]{22}}^{4}}{{\sqrt[5]{22}}^{1 + 4}}$

Add $1$ and $4$ .

$1 + r = \frac{\sqrt[5]{27} {\sqrt[5]{22}}^{4}}{{\sqrt[5]{22}}^{5}}$

Rewrite ${\sqrt[5]{22}}^{5}$ as $22$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[5]{22}$ as $22^{\frac{1}{5}}$ .

$1 + r = \frac{\sqrt[5]{27} {\sqrt[5]{22}}^{4}}{{(22^{\frac{1}{5}})}^{5}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$1 + r = \frac{\sqrt[5]{27} {\sqrt[5]{22}}^{4}}{22^{\frac{1}{5} \cdot 5}}$

Combine $\frac{1}{5}$ and $5$ .

$1 + r = \frac{\sqrt[5]{27} {\sqrt[5]{22}}^{4}}{22^{\frac{5}{5}}}$

Cancel the common factor of $5$ .

Cancel the common factor.

$1 + r = \frac{\sqrt[5]{27} {\sqrt[5]{22}}^{4}}{22^{\frac{5}{5}}}$

Rewrite the expression.

$1 + r = \frac{\sqrt[5]{27} {\sqrt[5]{22}}^{4}}{22^{1}}$

Evaluate the exponent.

$1 + r = \frac{\sqrt[5]{27} {\sqrt[5]{22}}^{4}}{22}$

Simplify the numerator.

Rewrite ${\sqrt[5]{22}}^{4}$ as $\sqrt[5]{22^{4}}$ .

$1 + r = \frac{\sqrt[5]{27} \sqrt[5]{22^{4}}}{22}$

Raise $22$ to the power of $4$ .

$1 + r = \frac{\sqrt[5]{27} \sqrt[5]{234256}}{22}$

Simplify the numerator.

Combine using the product rule for radicals.

$1 + r = \frac{\sqrt[5]{27 \cdot 234256}}{22}$

Multiply $27$ by $234256$ .

$1 + r = \frac{\sqrt[5]{6324912}}{22}$

Step 5

Subtract $1$ from both sides of the equation.

$r = \frac{\sqrt[5]{6324912}}{22} - 1$

Step 6

The result can be shown in multiple forms.

Exact Form:

$r = \frac{\sqrt[5]{6324912}}{22} - 1$

Decimal Form:

$r = 0.04180926 \dots$

Solve for r 2700=2200(1+r)^5

Related Questions

Support LCMGCF.com by disabling your adblocker.