Algebra help

Show that n = $\frac{log2000003}{log3}-1$ from $\frac{3(1-3^n)}{1-3} = 1000000$ without expanding the brackets. I'm not sure how to do this, thanks.

Is your denominator supposed to be 1-3 or something else?

You have to expand the brackets to get n on its own

It's supposed to be 1-3



Can't you multiply the equation by -2 and divide by 3 to get n on its own? Isn't this able to be done this way as well?

Weird way to write the denominator, but yes, multiplying by -2 and dividing by 3 and then using log laws should get you the right answer.

Yeah, sorry. It's a sum of a geometric sequence question.
I already tried doing what you said and I couldn't get anywhere.
I got to $1-3^n = -2000000/3$
bringing over the one and dividing by -1 gives 3^n = 2000000/3 +1

So 200003/3 = 3^n, taking log base 3 of each side gives log(3) 2000003/3 = n, and then using the log law about divisibility - log a/b = loga - logb so log(3)20000003 -log(3)3 = n, and then there's the log law of log(a)a = 1 which is where your -1 comes from, so you end up with log(3)200003 -1 =n.

Hope that's not too bad to read, I can't be doing with latex at this time.

But the question wants it in terms of base 10 and the log of 2000003 should be divided by the log of 3. It's fine, i'll just remember to expand the brackets every time i see a log question. This will bug me forever, though. Thanks alot.

Ah sorry, thats me misreading the question. Take base ten then, so you've got 2000003/3 = 3^n, and then log(10)200003/3 = log(10)3^n, then using the division log law thing again you get log 2000003 - log 3 = log3^n, taking the log 3 to the right hand side and using the multiplying log law you get log 200003 = (n+1)log3 (and using the log law where you bring the powers down) and then divide by log 3, take across the 1 and you get the answer, just with a bit more fiddling about then using base 3!