Use logarithms to evaluate sqrt((14.70 35^ 24')/(69.852)).

Step 1: Express the given expression in logarithmic form.
Let $X = \sqrt{\frac{14.70 \tan 35^\circ 24'}{69.85^2}}$.
First, convert $35^\circ 24'$ to decimal degrees:

$$35^\circ 24' = 35^\circ + \frac{24}{60}^\circ = 35^\circ + 0.4^\circ = 35.4^\circ$$

So, the expression becomes:

$$X = \left(\frac{14.70 \tan 35.4^\circ}{69.85^2}\right)^{1/2}$$

Taking the base-10 logarithm of both sides:

$$\log X = \log \left[ \left(\frac{14.70 \tan 35.4^\circ}{69.85^2}\right)^{1/2} \right]$$

Using the logarithm property $\log(a^b) = b \log a$:

$$\log X = \frac{1}{2} \log \left(\frac{14.70 \tan 35.4^\circ}{69.85^2}\right)$$

Using the logarithm properties $\log\left(\frac{a}{b}\right) = \log a - \log b$ and $\log(ab) = \log a + \log b$:

$$\log X = \frac{1}{2} [\log(14.70) + \log(\tan 35.4^\circ) - \log(69.85^2)]$$

Using the logarithm property $\log(a^b) = b \log a$ again:

$$\log X = \frac{1}{2} [\log(14.70) + \log(\tan 35.4^\circ) - 2 \log(69.85)]$$

Step 2: Calculate the logarithms of the individual terms.
Using a calculator for base-10 logarithms:

$$\log(14.70) \approx 1.1673173$$

First, calculate $\tan(35.4^\circ)$:

$$\tan(35.4^\circ) \approx 0.7107090$$

Then, calculate its logarithm:

$$\log(\tan 35.4^\circ) = \log(0.7107090) \approx -0.1483864$$ $$\log(69.85) \approx 1.8441530$$

Step 3: Substitute these values into the equation for $\log X$.

$$\log X = \frac{1}{2} [1.1673173 + (-0.1483864) - 2(1.8441530)]$$ $$\log X = \frac{1}{2} [1.1673173 - 0.1483864 - 3.6883060]$$ $$$$