b) $$ \log x+\log 20-\log 3=2 $$

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1. Combine the logarithms using the properties of logarithms. Recall that
   $$
   \log a + \log b = \log(ab) \quad \text{and} \quad \log a - \log b = \log\left(\frac{a}{b}\right).
   $$
   Therefore,
   $$
   \log x + \log 20 - \log 3 = \log\left(\frac{20x}{3}\right).
   $$

2. Rewrite the equation:
   $$
   \log\left(\frac{20x}{3}\right) = 2.
   $$

3. Convert the logarithmic equation to its equivalent exponential form. Using the definition of the logarithm, where $$\log_{10} y = z$$ implies $$y = 10^z$$, we have:
   $$
   \frac{20x}{3} = 10^2.
   $$

4. Compute $$10^2$$:
   $$
   10^2 = 100,
   $$
   so the equation becomes:
   $$
   \frac{20x}{3} = 100.
   $$

5. Solve for $$x$$. Multiply both sides by 3:
   $$
   20x = 100 \times 3 = 300.
   $$
   Then divide by 20:
   $$
   x = \frac{300}{20} = 15.
   $$

The solution is $$ x = 15 $$.
$$ x = 15 $$

b) \( \log x+\log 20-\log 3=2 \)

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