Solve 2520/2400=[1+8/100]^4/[1+frac{8{100}]^3}

$$\frac { 2520 } { 2400 } = \frac { [ 1 + \frac { 8 } { 100 } ] ^ { 4 } } { [ 1 + \frac { 8 } { 100 } ] ^ { 3 } }$$

Verify

Solution Steps

To divide powers of the same base, subtract the denominator's exponent from the numerator's exponent. Subtract 3 from 4 to get 1.

$$\frac{2520}{2400}=\left(1+\frac{8}{100}\right)^{1}$$

Reduce the fraction \frac{2520}{2400} to lowest terms by extracting and canceling out 120.

$$\frac{21}{20}=\left(1+\frac{8}{100}\right)^{1}$$

Reduce the fraction \frac{8}{100} to lowest terms by extracting and canceling out 4.

$$\frac{21}{20}=\left(1+\frac{2}{25}\right)^{1}$$

Add 1 and \frac{2}{25} to get \frac{27}{25}.

$$\frac{21}{20}=\left(\frac{27}{25}\right)^{1}$$

Calculate \frac{27}{25} to the power of 1 and get \frac{27}{25}.

$$\frac{21}{20}=\frac{27}{25}$$

Least common multiple of 20 and 25 is 100. Convert \frac{21}{20} and \frac{27}{25} to fractions with denominator 100.

$$\frac{105}{100}=\frac{108}{100}$$

Compare \frac{105}{100} and \frac{108}{100}.

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