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The answer is 16
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How do you solve the equation 5.2 log4 2x16?
Due to limitations with browsers mathematical operators (especially + =) get stripped from questions (leaving questions with not enough information to answer them) and it is not entirely clear what the log4 bit means. I guess that the log4 bit is logarithms to base 4 of 2x^16 (which I'll write as log_4(2x^16) for brevity). If this is so, use normal algebraic operations to make log_4(2x^16) the subject of the equation. With logs there are useful rules; given 2 numbers 'a' and 'b': log(ab) = log(a) + log(b) log(a^b) = b × log(a) Which means: log_4(2x^16) = log_4(2) + log_4(x^16) = log_4(2) + 16 × log(x) and the equation can be further rearranged: log_4(2x^16) = <whatever> → log_4(2) + 16 × log(x) = <whatever> → log(x) = (<whatever> - log_4(2)) / 16 Logarithms tell you the power to which the base of the logarithm must be raised to get its argument, for example when using common logs: lg 100 = 2 since 10 must be raised to the power 2 to get 100, ie 10² = 100. (lg is the abbreviation for logs to base 10; ln, or natural logs, is the abbreviation for logs to the base e.) With logs to base 4, it is 4 that is raised to the power of the log to get the original value. eg log_4(16) = 2 since 4^2 = 16. log_4(2) can be worked out: The log to any base of the base is 1 (since any number to the power 1 is itself). Now 2 × 2 = 2² = 4. → log_4(4) = 1 → log_4(2²) = 1 → 2 × log_4(2) = 1 → log_4(2) = ½ → log(x) = (<whatever> - ½) / 16 Back to the rearranged equation; with logs to base 4, if you make both sides the power of 4 you'll get: 4^(log_4(x)) = 4^(<whatever>) → x = 4^(<whatever>) which now solves for x.
Evaluate 81 to the power of log 16 with a base of 9?
It is 256.
How do you simplify log base 3 25 plus log base 3 4 and express using base 10 logarithms?
log325 + log34 = log3(25*4) = log3(100) = log10100/log103 = 2/log103
How do you key in log 2 on your TI 86 calculator?
Be careful . On calculatoirs there are TWO logarithm bases, indicated by 'log' and 'ln'. They are not interchangeable. 'log' is logs to base '10' 'ln' is logs to the 'natural' base ; natural = 2.718281828.... Try 'log' , 'number'. '=' and the answer should appear. e.g. log(4) = 0.6020599999.... ln(4) = 1.386294371.... Note the two different answers. Notwithstanding, what is written above, by a special higher level mathemtics , log bases can be changed. However, whilst learning logarithms, keep to 'base 10' ( log).
What is the same thing as log base e?
log base e = ln.
Which logarithm is equivalent to log base 3 16 - log base 3 2?
log316 - log32 = log38
How do you solve log base16 8?
You divide log 8 / log 16. Calculate the logarithm in any base, but use the same base for both - for example, ln 8 / ln 16.
Log base 3 of 81 plus log base 3 of 81?
Log base 3 of 81 is equal to 4, because 3 ^ 4 = 81. Therefore, two times log base 3 of 81 is equal to 2 x 4 = 8.
How do you solve the equation 5.2 log4 2x16?
Due to limitations with browsers mathematical operators (especially + =) get stripped from questions (leaving questions with not enough information to answer them) and it is not entirely clear what the log4 bit means. I guess that the log4 bit is logarithms to base 4 of 2x^16 (which I'll write as log_4(2x^16) for brevity). If this is so, use normal algebraic operations to make log_4(2x^16) the subject of the equation. With logs there are useful rules; given 2 numbers 'a' and 'b': log(ab) = log(a) + log(b) log(a^b) = b × log(a) Which means: log_4(2x^16) = log_4(2) + log_4(x^16) = log_4(2) + 16 × log(x) and the equation can be further rearranged: log_4(2x^16) = <whatever> → log_4(2) + 16 × log(x) = <whatever> → log(x) = (<whatever> - log_4(2)) / 16 Logarithms tell you the power to which the base of the logarithm must be raised to get its argument, for example when using common logs: lg 100 = 2 since 10 must be raised to the power 2 to get 100, ie 10² = 100. (lg is the abbreviation for logs to base 10; ln, or natural logs, is the abbreviation for logs to the base e.) With logs to base 4, it is 4 that is raised to the power of the log to get the original value. eg log_4(16) = 2 since 4^2 = 16. log_4(2) can be worked out: The log to any base of the base is 1 (since any number to the power 1 is itself). Now 2 × 2 = 2² = 4. → log_4(4) = 1 → log_4(2²) = 1 → 2 × log_4(2) = 1 → log_4(2) = ½ → log(x) = (<whatever> - ½) / 16 Back to the rearranged equation; with logs to base 4, if you make both sides the power of 4 you'll get: 4^(log_4(x)) = 4^(<whatever>) → x = 4^(<whatever>) which now solves for x.
What is log base 4 of 27?
You can calculate that on any scientific calculator - like the calculator on Windows (if you change the options, to display as a scientific calculator). Log base 4 of 27 is the same as log 27 / log 4. You can use logarithms in any base to calculate that - just use the same base for both logarithms.
Evaluate 81 to the power of log 16 with a base of 9?
It is 256.
What is log base 64 of 4?
If log644 = x, then 64x = 4. The cubed root of 64 (which is the same as 641/3) is 4, so log base 64 of 4 is 1/3.
121 in base 4 plus 84 in base 16?
121 in base 4 = 16 + 8 + 1 (base ten) = 2584 in base 16 = 128 + 4 (base ten) = 132132 + 25 = 157(base 10) = 9D (base 16) = 2131 (base 4)
What is logarithm in simple terms?
The power to which a 'base number' (usually 10) has to be raised to produce a given number. as an example: Log (base 10) of 100 = 2 ............ because 10 raised to the power of 2 (or 10 squared) or 10 x 10 = 100 log (base 10) of 1000 = 3 ........... because 10 raised to the power of 3 (or 10 cubed) or 10 x 10 x10 = 1000 log (base 10) of 1000000000 = 9 ... because 10 raised to the powr of 9 or 10x10x10x10x10x10x10x10x10 = 1000000000 In a similar way log (base 2) of 16 = 4................. because 2x2x2x2 (2 raised to the power of 4) = 16 and so on.
How do you solve 16 as a power of 2?
In this case, trial and error is probably the easiest: 22 = 4, 23 = 8, 24 = 16 yes! A more genral answer is: the power that you want is log(16)/log(2) where the logarithm is calculated to any base (10 or e , or indeed any other).
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 you simplify log base 3 25 plus log base 3 4 and express using base 10 logarithms?
log325 + log34 = log3(25*4) = log3(100) = log10100/log103 = 2/log103
