Solve for r 2700=2300(1+r)^5

$2700 = 2300 {(1 + r)}^{5}$

Step 1

Rewrite the equation as $2300 {(1 + r)}^{5} = 2700$ .

$2300 {(1 + r)}^{5} = 2700$

Step 2

Divide each term in $2300 {(1 + r)}^{5} = 2700$ by $2300$ and simplify.

Divide each term in $2300 {(1 + r)}^{5} = 2700$ by $2300$ .

$\frac{2300 {(1 + r)}^{5}}{2300} = \frac{2700}{2300}$

Simplify the left side.

Cancel the common factor of $2300$ .

Cancel the common factor.

$\frac{2300 {(1 + r)}^{5}}{2300} = \frac{2700}{2300}$

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

${(1 + r)}^{5} = \frac{2700}{2300}$

Simplify the right side.

Cancel the common factor of $2700$ and $2300$ .

Factor $100$ out of $2700$ .

${(1 + r)}^{5} = \frac{100 (27)}{2300}$

Cancel the common factors.

Factor $100$ out of $2300$ .

${(1 + r)}^{5} = \frac{100 \cdot 27}{100 \cdot 23}$

Cancel the common factor.

${(1 + r)}^{5} = \frac{100 \cdot 27}{100 \cdot 23}$

Rewrite the expression.

${(1 + r)}^{5} = \frac{27}{23}$

Step 3

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

$1 + r = \sqrt[5]{\frac{27}{23}}$

Step 4

Simplify $\sqrt[5]{\frac{27}{23}}$ .

Rewrite $\sqrt[5]{\frac{27}{23}}$ as $\frac{\sqrt[5]{27}}{\sqrt[5]{23}}$ .

$1 + r = \frac{\sqrt[5]{27}}{\sqrt[5]{23}}$

Multiply $\frac{\sqrt[5]{27}}{\sqrt[5]{23}}$ by $\frac{{\sqrt[5]{23}}^{4}}{{\sqrt[5]{23}}^{4}}$ .

$1 + r = \frac{\sqrt[5]{27}}{\sqrt[5]{23}} \cdot \frac{{\sqrt[5]{23}}^{4}}{{\sqrt[5]{23}}^{4}}$

Combine and simplify the denominator.

Multiply $\frac{\sqrt[5]{27}}{\sqrt[5]{23}}$ by $\frac{{\sqrt[5]{23}}^{4}}{{\sqrt[5]{23}}^{4}}$ .

$1 + r = \frac{\sqrt[5]{27} {\sqrt[5]{23}}^{4}}{\sqrt[5]{23} {\sqrt[5]{23}}^{4}}$

Raise $\sqrt[5]{23}$ to the power of $1$ .

$1 + r = \frac{\sqrt[5]{27} {\sqrt[5]{23}}^{4}}{{\sqrt[5]{23}}^{1} {\sqrt[5]{23}}^{4}}$

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

$1 + r = \frac{\sqrt[5]{27} {\sqrt[5]{23}}^{4}}{{\sqrt[5]{23}}^{1 + 4}}$

Add $1$ and $4$ .

$1 + r = \frac{\sqrt[5]{27} {\sqrt[5]{23}}^{4}}{{\sqrt[5]{23}}^{5}}$

Rewrite ${\sqrt[5]{23}}^{5}$ as $23$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[5]{23}$ as $23^{\frac{1}{5}}$ .

$1 + r = \frac{\sqrt[5]{27} {\sqrt[5]{23}}^{4}}{{(23^{\frac{1}{5}})}^{5}}$

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

$1 + r = \frac{\sqrt[5]{27} {\sqrt[5]{23}}^{4}}{23^{\frac{1}{5} \cdot 5}}$

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

$1 + r = \frac{\sqrt[5]{27} {\sqrt[5]{23}}^{4}}{23^{\frac{5}{5}}}$

Cancel the common factor of $5$ .

Cancel the common factor.

$1 + r = \frac{\sqrt[5]{27} {\sqrt[5]{23}}^{4}}{23^{\frac{5}{5}}}$

Rewrite the expression.

$1 + r = \frac{\sqrt[5]{27} {\sqrt[5]{23}}^{4}}{23^{1}}$

Evaluate the exponent.

$1 + r = \frac{\sqrt[5]{27} {\sqrt[5]{23}}^{4}}{23}$

Simplify the numerator.

Rewrite ${\sqrt[5]{23}}^{4}$ as $\sqrt[5]{23^{4}}$ .

$1 + r = \frac{\sqrt[5]{27} \sqrt[5]{23^{4}}}{23}$

Raise $23$ to the power of $4$ .

$1 + r = \frac{\sqrt[5]{27} \sqrt[5]{279841}}{23}$

Simplify the numerator.

Combine using the product rule for radicals.

$1 + r = \frac{\sqrt[5]{27 \cdot 279841}}{23}$

Multiply $27$ by $279841$ .

$1 + r = \frac{\sqrt[5]{7555707}}{23}$

Step 5

Subtract $1$ from both sides of the equation.

$r = \frac{\sqrt[5]{7555707}}{23} - 1$

Step 6

The result can be shown in multiple forms.

Exact Form:

$r = \frac{\sqrt[5]{7555707}}{23} - 1$

Decimal Form:

$r = 0.03258826 \dots$

Solve for r 2700=2300(1+r)^5

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