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What is log base 4 of 16?
The answer is 16
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 2 to the power of 4?
What is 1+2 with sum
How do you evaluate 4 to the third power?
4^3 = 4*4*4 = 16*4 = 64
Can you give an example of an evaluation question in math?
Evaluate: 4x + 16 = 32 4x + 16 -16 = 32 - 16 4x = 16 x = 4 evaluate basically means manipulate and find the answer
What is log base 4 of 16?
The answer is 16
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.
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 2 to the power of 4?
What is 1+2 with sum
How do you evaluate 4 to the 2nd power?
42 = 4*4 = 16
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).
What is logarithm in simple terms?
A logarithm is the inverse operation of exponentiation. It is used to find the power to which a fixed number (called the base) must be raised to produce a given number. Logarithms help simplify calculations involving very large or very small numbers.
How do you evaluate 4 to the third power?
4^3 = 4*4*4 = 16*4 = 64
How do you write a number as a power of 2?
For an exact power of 2 (1, 2, 4, 8, 16, 32, etc., but also 1/2, 1/4, etc.), you can try out different exponents until you get it right. To write any number (greater than 0) as a power of 2 is equivalent to taking the logarithm of that number in base 2, which is the same (if you call your number "n") as calculating log n / log 2 (using the same base for both logarithms - for example, both in base 10, or both in base e).
Can you give an example of an evaluation question in math?
Evaluate: 4x + 16 = 32 4x + 16 -16 = 32 - 16 4x = 16 x = 4 evaluate basically means manipulate and find the answer
Write each number as power of given base 16 base2?
4
