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Twelfth grader Monesa from Pakistan writes, "A geometric sequence has first term 1/9 and common ratio 3. (i) Find the fifth term. (ii) Which is the first term of the sequence which exceeds 1000? I've found (i) as 9, but cannot do (ii) without trial and error. I think it may have something to do with logarithms, but can't think how to do it. Any help would be appreciated"
Hi Monesa, you absolutely can use logarithms to solve the second part of your question. Here's how that looks. First, we write a non-logarithmic equation just like you would for any geometric sequence problem.
1/9 = 1000.
Strictly speaking, we want the term to be greater than or equal to 1000, but for now we're going to write it as an equation; this will give us the minimum possible value of n.
Multiplying both sides of the equation by 9 gives:
3n - 1 = 9000.
Here's where logarithms come into play. If you've studied logarithms at all, you know that a log is actually just an exponent. You also have been taught that:
ab = c and loga c = b are equivalent equations. So we can take your exponential function above and rewrite it as a log equation:
log 3 9000 = n - 1
Now, your calculator probably won't calculate a log in base 3, which is what you have here. But we have a nice log property called the "Change of Base Property" - it says:
log a b= (log c b)/(log c a)
In other words, we can change the base to anything we want. What base do we want? Base ten, because our calculators will handle base ten logarithms:
log 3 9000 = (log 9000)/(log 3) = 8.29...
Wait a minute! That gives n = 9.29! That doesn't make sense; how can you have term number 9.29? The answer is, you can't. In a geometric sequence, the term number has to be an integer. But we know that it has to be at least 9.29, so the smallest integer value n could be is 10. The tenth term is larger than 1000.
Can we check that? Sure! 1/9 · 39 = 2187, and 1/9 · 38 = 729. Sure enough, the 9th term doesn't quite make it to 1000.