Which term of the series 8 - 4 + 2 - ... is equal to 1/128?
1/128 = 8 x (-1/2)^(n-1)
1/1024 = (-1/2)^(n-1)
log(1/1024) = log (-1/2)^(n-1)
log(1/1024) = (n-1) log (-1/2)
n-1 = log(1/1024) / log (-1/2)
and then my calculator gives me an error. Help?
It's a very simple question and I'm usually good at maths but this has stumped me, what am I doing wrong?
By the way, we haven't done the Logs topic yet so I don't know any of the rules. Our teacher just showed us how to do this to solve geometric series, which is our current topic.
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M$2 Answers
The logarithm applies only to positive numbers
Look closely at this
1/1024=(-1/2)^(n-1)
The left hand is positive, so the right hand should be also positive
Now, if n-1 is odd, then (-1/2)^(n-1) would be negative
if n-1 is even, then (-1/2)^(n-1) would be positive
Therefore n-1 should be even.
In this case, the negative sign in the bracket disappears because
(-1/2)^(n-1)=(1/2)^(n-1) when n-1 is even
So now you have
1/1024=(1/2)^(n-1)
Can you continue from here now?
Also can you try to solve this without logarithms(so without calculator)? Hint: what power of 2 is 128 and 1024?
if you still have trouble with why the negative sign disappears when the power is even, think about
(-10)^2=10^2=100
but (-10)^3=-1000 differs than 10^3=1000
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M$
ALL EXPONENTS ARE SHOWN IN BLUE. * DENOTES MULTIPLY.
1 / 128 = 8 * ( -1 / 2 ) ^ N
2 ^ -7 = 2 ^ 3 * ( 2 ^ -1 ) ^ N
2 ^ 7 = 2 ^ 3 * ( 2 ^ -N )
2 ^ 7 = 2 ^ 3 - N
-7 = 3 - N
N = 3 + 7
N = 10
Conclusion: The 10th term in the geometric series is 1 / 128.
Check: 1 / 128 = 8 * ( -1 / 2 ) ^ 10 = 1 / 128
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M$
I see. Thanks. I did the question using the powers of 2, which was easy.
The question works out when I use positive 1/2, and I get the right answer.. I'm just wondering, is that a law of logarithms?
I tipped you M1$ for this awesome answer.
Thank you-I appreciate it
Yes, the logarithm is defined only for positive numbers
Look at this image: the logarithm (in base e) graph is below the green line and it is evaluated only at positive numbers x.
The x-axis is as usual the horizontal line
we can say that the domain of the logarithmic function
log_b(x) is (0,∞)