To solve for $x$ in the equation:

$\log(9x - 7) = 1$

we need to rewrite it in exponential form. Since the logarithm has a base of 10 (implied in $\log$), we can rewrite this as:

$10^1 = 9x - 7$

Simplify this:

$10 = 9x - 7$

Next, add 7 to both sides:

$10 + 7 = 9x$

$17 = 9x$

Now, divide by 9:

$x = \frac{17}{9}$

So, the exact answer is:

$x = \frac{17}{9}$

Would you like further explanation on any step?

Here are some related questions to explore further:

1. How would the answer change if the equation was $\log(9x - 7) = 2$?
2. What if the logarithm had a different base, like $\log_2(9x - 7) = 1$?
3. Can you explain the properties of logarithms that allow converting between logarithmic and exponential forms?
4. How would you solve $\ln(5x + 3) = 2$?
5. How would you verify the solution to this equation?

Tip: When converting logarithmic equations to exponential form, remember that $\log_b(y) = x$ translates to $b^x = y$.