didn't take too long the way I did it, hurrried and multiplied everything out from 3^2 to 3^5..... 3^5 gives us 243, while 3^4 gives us 81, subtract the two and you get 162...... but we're concerened with the x's, so.... 5(5-1)=20

The left side of the equation can be factored... here's how it would unfold:

3^x - 3^(x-1) = 162
3^(x-1) [3 - 1] = 162 ( factored out a 3^(x-1) )
3^(x-1) = 81
3^(x-1) = 3^4
x = 5

x(x-1) = 20

Thanks. Can you explain that factoring? I get lost with the 3^x-1 of a factor

Sure. I think it's easier to see it with a similar example.

Suppose you want to factor n^6 - n^5 buy "pulling out" the common factor. Well, in this case the common factor would be the
smallest power of n that is common to both terms, namely n^5. So, you pull that out. What remains in the parentheses are the original terms divided by n^5.

n^6 - n^5
n^5(n - 1)

Notice that we would get the original expression by distributing back through. With these tricky factoring problems, I always perform a quick mental check by distributing through.

In the problem, we have

3^x - 3^(x-1)

The smallest power of 3 between the two terms is (x-1), so we pull out 3^(x-1). The tricky part is figuring out what belongs in the parentheses. Well, just as above, we just need to divide each term by 3^(x-1). Alternatively, you could think of what needs to be in the parentheses so that if we were to distribute through again, we'd wind up with the original expression.

By either reasoning, 3^x yields 3^1 or just 3. 3^(x-1) yields 1 since it is the common factor.

3^(x-1) [3 - 1]
Notice that we can distribute back through and get 3^x - 3^(x-1).

Then you simplify:
3^(x-1)*2

Rey