There are 3 number a, b and c such that ` log_(10) a = 5.71, log_(10) b = 6.23 and log_(10) c = 7.89`. Find the number of digits before dicimal in ` (ab^(2))/c`.

Explanation

To find the number of digits before the decimal in the expression \((ab^2)/c\), we can follow these steps:

### Step 1: Define the expression
Let \( n = \frac{ab^2}{c} \).

### Step 2: Take the logarithm
Taking the logarithm (base 10) of both sides, we have:
\[ \log_{10} n = \log_{10} \left( \frac{ab^2}{c} \right) \]

### Step 3: Apply logarithmic properties
Using the properties of logarithms, we can rewrite the expression:
\[ \log_{10} n = \log_{10} (ab^2) - \log_{10} c \]
Using the property \(\log_{10} (xy) = \log_{10} x + \log_{10} y\) and \(\log_{10} (x^y) = y \log_{10} x\), we get:
\[ \log_{10} n = \log_{10} a + \log_{10} (b^2) - \log_{10} c = \log_{10} a + 2 \log_{10} b - \log_{10} c \]

### Step 4: Substitute the given values
We know:
- \(\log_{10} a = 5.71\)
- \(\log_{10} b = 6.23\)
- \(\log_{10} c = 7.89\)

Substituting these values into the equation:
\[ \log_{10} n = 5.71 + 2(6.23) - 7.89 \]

### Step 5: Calculate the logarithm
Now, calculate \(2 \log_{10} b\):
\[ 2 \log_{10} b = 2 \times 6.23 = 12.46 \]
Now substitute this back into the equation:
\[ \log_{10} n = 5.71 + 12.46 - 7.89 \]

### Step 6: Simplify the expression
Now, perform the addition and subtraction:
\[ \log_{10} n = 5.71 + 12.46 = 18.17 \]
\[ \log_{10} n = 18.17 - 7.89 = 10.28 \]

### Step 7: Determine the number of digits
The number of digits before the decimal point in a number \( n \) can be found using the characteristic of the logarithm:
- If \(\log_{10} n = x\), then the number of digits before the decimal point is given by \(\lfloor x \rfloor + 1\).

In our case:
\[ \lfloor 10.28 \rfloor + 1 = 10 + 1 = 11 \]

### Final Answer
Thus, the number of digits before the decimal in \((ab^2)/c\) is **11**.

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