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log2(31000) = 1000 log2(3)

log2(3) = 1.585 (rounded)

1000 log2(3) = log2(31000) = 1,584.96(rounded)

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Q: How do you solve log base 2 of 3 to the 1000?
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How do you solve log base 2 of x - 3 log base 2 of 5 equals 2 log base 2 of 10?

[log2 (x - 3)](log2 5) = 2log2 10 log2 (x - 3) = 2log2 10/log2 5 log2 (x - 3) = 2(log 10/log 2)/(log5/log 2) log2 (x - 3) = 2(log 10/log 5) log2 (x - 3) = 2(1/log 5) log2 (x - 3) = 2/log 5 x - 3 = 22/log x = 3 + 22/log 5


What is log 100 base e?

log 100 base e = log 100 base 10 / log e base 10 log 100 base 10 = 10g 10^2 base 10 = 2 log 10 base 10 = 2 log e base 10 = 0.434294 (calculator) log 100 base e = 2/0.434294 = 4.605175


Log base 2 of x - log base 2 of x-23?

log base 2 of [x/(x - 23)]


Suppose aneq 0. Compute log 2a 2b in terms of a and b.?

Due to the rubbish browser that we are compelled to use, it is not possible to use any super or subscripts so here goes, with things spelled out in detail: log to base 2a of 2b = log to base a of 2b/log to base a of 2a = [(log to base a of 2) + (log to base a of b)] / [(log to base a of 2) + (log to base a of a)] = [(log to base a of 2) + (log to base a of b)] / [(log to base a of 2) + 1]


How do I solve the equation e to the power of x equals 2 using logarithms?

When the logarithm is taken of any number to a power the result is that power times the log of the number; so taking logs of both sides gives: e^x = 2 → log(e^x) = log 2 → x log e = log 2 Dividing both sides by log e gives: x = (log 2)/(log e) The value of the logarithm of the base when taken to that base is 1. The logarithms can be taken to any base you like, however, if the base is e (natural logs, written as ln), then ln e = 1 which gives x = (ln 2)/1 = ln 2 This is in fact the definition of a logarithm: the logarithm to a specific base of a number is the power of the base which equals that number. In this case ln 2 is the number x such that e^x = 2. ---------------------------------------------------- This also means that you can calculate logs to any base if you can find logs to a specific base: log (b^x) = y → x log b = log y → x = (log y)/(log b) In other words, the log of a number to a given base, is the log of that number using any [second] base you like divided by the log of the base to the same [second] base. eg log₂ 8 = ln 8 / ln 2 = 2.7094... / 0.6931... = 3 since log₂ 8 = 3 it means 2³ = 8 (which is true).


How do you solve log x - 2?

You cannot solve log x- 2 unless (i) log x - 2 is equal to some number or (ii) x is equal to some number.


What is the value of x if log base 10 of x equals 2?

Logs to base 10 are common logs and are abbreviated to lg; to solve use antilogs: lg x = 2 → 10^(lg x) = 10^2 → x = 10² = 100


What is logarithm in simple terms?

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a log is the 'undo-er' of powers, kind of like division is the 'undo-er' of multiplication. EX: 102 = 100, then log10(100) = 2 103 = 1000, then log10(1000) = 3, in this example, we are using log base 10, this is a default base and sometimes isn't even wirten. e is probably the most common base but log base e is more simply called the natural log, or ln. so in general: logx(m) = N means that xN = m so log5(125) = 3 because 53 = 125.


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inverse log of 2= 1/(log{10}2)= 1/(log2)=1/0.3010299=3.3219. hence answer is 3.3219


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