A) Show that log(70/33) + log(22/135) - log(7/18) = 3log2 − 2log3. b) If log10 x = a, compute the value of 10^(a-1) in terms of x.

$$\text{LHS} = \log \left( \frac{70}{33} \times \frac{22}{135} \right) - \log \frac{7}{18}$$

First, simplify the product inside the logarithm:

$$\frac{70}{33} \times \frac{22}{135} = \frac{70 \times 22}{33 \times 135}$$

Factorize the numbers to simplify:

$$\frac{(2 \times 5 \times 7) \times (2 \times 11)}{(3 \times 11) \times (5 \times 3^3)} = \frac{2^2 \times 5 \times 7 \times 11}{3^4 \times 5 \times 11}$$

Cancel common factors (5 and 11):

$$= \frac{2^2 \times 7}{3^4} = \frac{4 \times 7}{81} = \frac{28}{81}$$

Now substitute this back into the LHS:

$$\text{LHS} = \log \left( \frac{28}{81} \right) - \log \left( \frac{7}{18} \right)$$

Apply the subtraction property of logarithms:

$$\text{LHS} = \log \left( \frac{28/81}{7/18} \right) = \log \left( \frac{28}{81} \times \frac{18}{7} \right)$$

Simplify the fraction:

$$\frac{28}{81} \times \frac{18}{7} = \frac{(4 \times 7) \times (2 \times 9)}{(9 \times 9) \times 7}$$

Cancel common factors (7 and 9):

$$= \frac{4 \times 2}{9} = \frac{8}{9}$$

So, the LHS simplifies to:

$$\text{LHS} = \log \left( \frac{8}{9} \right)$$

Rewrite $\frac{8}{9}$ using powers of 2 and 3:

$$\frac{8}{9} = \frac{2^3}{3^2}$$

Substitute this into the LHS:

$$\text{LHS} = \log \left( \frac{2^3}{3^2} \right)$$

Apply the logarithm properties $\log\left(\frac{A}{B}\right) = \log A - \log B$ and $\log(A^n) = n \log A$:

$$\text{LHS} = \log(2^3) - \log(3^2)$$ $$\text{LHS} = 3 \log 2 - 2 \log 3$$

This matches the right-hand side (RHS) of the given equation.
Therefore, the statement is shown.

Given $\log_{10} x = a$.
By the definition of a logarithm, this means:

$$10^a = x$$

Recall the exponent rule $A^{B-C} = \frac{A^B}{A^C}$.

$$10^{a-1} = \frac{10^a}{10^1}$$

Substitute $10^a = x$ into the expression:

$$10^{a-1} = \frac{x}{10}$$

The value of $10^{a-1}$ in terms of $x$ is $\boxed{\frac{x}{10}}$.