What is the answer to 2^r-(1+r/(100))^{72}=0 ?

The answer to 2^r-(1+r/(100))^{72}=0 is

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2^r-(1+r/(100))^{72}=0

2^{r} - (1 + \frac{r}{100})^{72} = 0

r = - \frac{72 W _{0} ( \frac{25 l n ( 2 )}{18 e ^{\frac{25 l n ( 2 )}{18}}} )}{ln ( 2 )} - 100, r = - \frac{72 W _{0} ( - \frac{25 l n ( 2 )}{18 e ^{\frac{25 l n ( 2 )}{18}}} )}{ln ( 2 )} - 100, r = - \frac{72 W _{- 1} ( - \frac{25 l n ( 2 )}{18 e ^{\frac{25 l n ( 2 )}{18}}} )}{ln ( 2 )} - 100

Decimal Notation

r = - 128.90910 \dots, r = 0, r = 7.84687 \dots

Solution steps

2^{r} - (1 + \frac{r}{100})^{72} = 0

Move

(1 + \frac{r}{100})^{72}

to the right side

2^{r} = (\frac{r}{100} + 1)^{72}

Take both sides of the equation to the power of

\frac{1}{72}

(2^{r})^{\frac{1}{72}} = ((\frac{r}{100} + 1)^{72})^{\frac{1}{72}}

Simplify

(2^{r})^{\frac{1}{72}} : 2^{\frac{r}{72}}

Solve

2^{\frac{r}{72}} = \frac{r}{100} + 1 : r = - \frac{72 W _{0} ( \frac{25 l n ( 2 )}{18 e ^{\frac{25 l n ( 2 )}{18}}} )}{ln ( 2 )} - 100 or r = - \frac{72 W _{0} ( - \frac{25 l n ( 2 )}{18 e ^{\frac{25 l n ( 2 )}{18}}} )}{ln ( 2 )} - 100 or r = - \frac{72 W _{- 1} ( - \frac{25 l n ( 2 )}{18 e ^{\frac{25 l n ( 2 )}{18}}} )}{ln ( 2 )} - 100

r = - \frac{72 W _{0} ( \frac{25 l n ( 2 )}{18 e ^{\frac{25 l n ( 2 )}{18}}} )}{ln ( 2 )} - 100, r = - \frac{72 W _{0} ( - \frac{25 l n ( 2 )}{18 e ^{\frac{25 l n ( 2 )}{18}}} )}{ln ( 2 )} - 100, r = - \frac{72 W _{- 1} ( - \frac{25 l n ( 2 )}{18 e ^{\frac{25 l n ( 2 )}{18}}} )}{ln ( 2 )} - 100

\frac{2}{3 - 2 ^{x}} = 5

2 e^{2 x} = e^{x + 1}

a, f (a + 2) = 2^{(a + 2)}

5^{x + 1} + 6 \cdot 5^{x} = 6

2^{2 x} = 31

What is the answer to 2^r-(1+r/(100))^{72}=0 ?
The answer to 2^r-(1+r/(100))^{72}=0 is
Solve for r: 2^r-(1+r/(100))^{72}=0
The solution is