Question

ex.12 estimate \log_{10}(50)

Explanation:

Identify bounding powers of ten

To estimate \(\log_{10}(50)\), we find the consecutive powers of the base \(10\) that bracket the number \(50\).
We know:

$$10^1 = 10$$

$$10^2 = 100$$

Since \(10 < 50 < 100\), the value of \(\log_{10}(50)\) must lie strictly between \(1\) and \(2\).

Apply logarithmic properties

We can rewrite the expression using the product rule of logarithms to get a more precise estimate.

$$\log_{10}(50) = \log_{10}(10 \times 5)$$

Using the product rule \(\log_b(xy) = \log_b(x) + \log_b(y)\):

$$\log_{10}(50) = \log_{10}(10) + \log_{10}(5)$$

Since \(\log_{10}(10) = 1\):

$$\log_{10}(50) = 1 + \log_{10}(5)$$

Estimate the fractional part

We estimate \(\log_{10}(5)\) by comparing \(5\) to the powers of \(10\).
Since \(5\) is the geometric mean of \(1\) and \(25\), or we can approximate using known values:

• \(\log_{10}(1) = 0\)
• \(\log_{10}(10) = 1\)

Since \(5\) is halfway between \(1\) and \(10\) on a linear scale, its logarithm on a logarithmic scale is approximately \(0.7\) (since \(10^{0.7} \approx 5.01\)).
Thus:

$$\log_{10}(5) \approx 0.7$$

Calculate final estimate

Combine the integer and fractional parts to find the total estimate.

$$\log_{10}(50) = 1 + \log_{10}(5) \approx 1 + 0.7 = 1.7$$

The actual value is approximately \(1.699\), so \(1.7\) is an excellent estimate.

Answer:

Approximately \(1.7\) (or any value between \(1.6\) and \(1.7\))