How do I compute log(base 9)1/7?
I need to compute log base 9 (1/7). I understand that that means it's essentially 9^x=1/7, but I do not know how to determine what power of 9 results in 1/7. I know it's around 9^-.85, but that's not close enough. Please help.
Update:If it can be done on a calculator, please, let me know exactly how.
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You're question is stated well and I can tell that you've been thinking about it.
The first thing we can do is convert log_9 (1/7) to a form that we DO understand.
1/7 can be written as 7^(-1), so:
log_9 (1/7) = log_9 ( 7^-1)
The log will take that exponent away from the number inside and put it in front as a multiplicand:
log_9 ( 7^-1) = - log_9 (7)
In general, log_b (a) = log(a) / log(b)
So we have b = 9 and a = 7, so:
-log_9 (7) = -log(7) / log(9)
And the log functions on the right hand side are base 10, which we can evaluate with a calculator.
-log(7) / log(9) ~= -0.885621874580711130033964153541228859033235667297121739684...
-log(7) / log(9) ~= -0.88562187
And you'll find that 9^(-0.88562187) = 0.142857 which is very close to 1/7.
1/7 ~= 0.1428571428571428571428571428571428571428571428571428571428...
You can change the base using a standard formula. See:
http://www.ltcconline.net/greenl/Courses/154/logex...
Your understanding of logs is quite correct in that the logarithm is the power to which the base must be raised to generate the number, so what you want can be expressed as 1/(9 to the power of 9 that generates 7). That is 0.88562, close enough.
And 1/9^0.88562 = 1/7
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