The greatest number among `2^(60)`, `3^(48)`, `4^(36)` and `5^(24)` is

Options

A. `2^(60)`
B. `3^(48)`
C. `4^(36)`
D. `5^(24)`

Answer

To find the greatest number among \(2^{60}\), \(3^{48}\), \(4^{36}\), and \(5^{24}\), we can compare these numbers by expressing them in terms of a common base or by simplifying their powers.

### Step 1: Express all numbers in terms of powers of 2
We can rewrite \(4^{36}\) as follows:
\[ 4^{36} = (2^2)^{36} = 2^{72} \]

Now we have:
- \(2^{60}\)
- \(3^{48}\)
- \(2^{72}\) (from \(4^{36}\))
- \(5^{24}\)

### Step 2: Compare \(2^{60}\) and \(2^{72}\)
Since \(2^{72}\) is clearly greater than \(2^{60}\), we can eliminate \(2^{60}\) from our comparison. Now we need to compare \(3^{48}\) and \(5^{24}\) with \(2^{72}\).

### Step 3: Compare \(3^{48}\) and \(5^{24}\)
To compare \(3^{48}\) and \(5^{24}\), we can take the logarithm of both numbers to make the comparison easier.

Using the natural logarithm:
\[ \log(3^{48}) = 48 \log(3) \]
\[ \log(5^{24}) = 24 \log(5) \]

### Step 4: Calculate the values of \(48 \log(3)\) and \(24 \log(5)\)
Using approximate values:
- \(\log(3) \approx 0.477\)
- \(\log(5) \approx 0.699\)

Calculating:
\[ 48 \log(3) \approx 48 \times 0.477 = 22.896 \]
\[ 24 \log(5) \approx 24 \times 0.699 = 16.776 \]

### Step 5: Compare the results
Since \(22.896 > 16.776\), we conclude that:
\[ 3^{48} > 5^{24} \]

### Step 6: Compare \(3^{48}\) with \(2^{72}\)
Now we need to compare \(3^{48}\) with \(2^{72}\). We can also take logarithms:
\[ \log(2^{72}) = 72 \log(2) \]
Using \(\log(2) \approx 0.301\):
\[ 72 \log(2) \approx 72 \times 0.301 = 21.672 \]

Now we compare \(48 \log(3)\) with \(72 \log(2)\):
\[ 22.896 > 21.672 \]

### Conclusion
Since \(3^{48} > 2^{72}\) and \(3^{48} > 5^{24}\), we conclude that:
\[ 3^{48} \text{ is the greatest number among } 2^{60}, 3^{48}, 4^{36}, \text{ and } 5^{24}. \]

Thus, the greatest number is:
\[ \boxed{3^{48}} \]