What is 10000>5500(1+(0.3)/(100))^t ?

The solution to 10000>5500(1+(0.3)/(100))^t is t<log_{(100.3)/(100)}(20/11)

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10000>5500(1+(0.3)/(100))^t

10000 > 5500 (1 + \frac{0.3}{100})^{t}

t < lo g_{\frac{100.3}{100}} (\frac{20}{11})

Interval Notation

(- \infty, lo g_{\frac{100.3}{100}} (\frac{20}{11}))

Decimal Notation

t < 199.57776 \dots

Solution steps

10000 > 5500 (1 + \frac{0.3}{100})^{t}

Switch sides

5500 (1 + \frac{0.3}{100})^{t} < 10000

Divide both sides by

5500

(1 + \frac{0.3}{100})^{t} < \frac{20}{11}

Apply exponent rules

t < lo g_{\frac{100.3}{100}} (\frac{20}{11})

x^{1}

\frac{2 x}{3} - \frac{x + 1}{2} = 1

\frac{3 ( x - 1 )}{4} < 6

- 3 (x + 7) = 18

\frac{x + 2}{2} + 4 = \frac{x + 5}{7}

What is 10000>5500(1+(0.3)/(100))^t ?
The solution to 10000>5500(1+(0.3)/(100))^t is t<log_{(100.3)/(100)}(20/11)