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How do you solve k equals log4 91.8?
k=log4 91.8 4^k=91.8 -- b/c of log rules-- log 4^k=log 91.8 -- b/c of log rules-- k*log 4=log91.8 --> divide by log 4 k=log 91.8/log 4 k= 3.260
How do you solve log4 3-x -log2 x-1 0?
Unfortunately, the browser used by Answers.com for posting questions is incapable of accepting mathematical symbols. This means that we cannot see the mathematically critical parts of the question. We are, therefore unable to determine what exactly the question is about and so cannot give a proper answer to your question.In this question, for example, it is not clear whether the characters immediately after "log" are the bases. Also, it is unclear whether the first "term" is log(3-x) or log(3)-x: both logs to base 4.
What does log2x2 equal?
Since you did not provide a base for your logarithm there is no particular solution. In most cases it is best to assume it is to the base 10. So the answer would be: log(2*2)=log4 (to the base 10) log4=0.60205.... It is good practice to list the base to which the logarithm is at - normally written as a subscript in the lower right hand corner of the 'log'. The only exception is when the logarithm is to the base e, in which case we write it as ln(x) - where x is a real number. For more information check http://en.wikipedia.org/wiki/Logarithm
What is the expression with the logarithm Log 4 plus Log 3?
log4+log3=log(4x3)=log12
Log3 81 x log2 8 x log4 2 equals x?
log3 81 × log2 8 × log4 2 = log3 (33) × log2 (23) × log4 (40.5) = 3 × (log3 3) × 3 × (log2 2) × 0.5 × (log4 4) = 3 × 1 × 3 × 1 × 0.5 × 1 = 9 × 0.5 = 4.5
What is log4 19 to four right decimal places?
It is 2.1240
A logarithm is an exponent?
A logarithm is quite the opposite of an exponential function. Whereas an exponential is y=ax , a log is logay=x For example, log39=2 because you raise 3 by the 2nd power to get 9. In other words, log39=2 because 32=9 Logarithms are present because they are a handy way to solve exponential equations, and because calculators use them to great advantage.
The value of log log4log4x equals 1 then x equals?
the value of log (log4(log4x)))=1 then x=
Could you Condense the expression 5log4 2 plus 7 log4 x plus 4 log 4 y?
You use the identities: log(ab) = b log(a), and log(ab) = log a + log b. In this case, 5 log42 + 7 log4x + 4 log4y = log432 + log4x7 + log4y4 = log4 (32x7y4).
How do you solve k equals log4 91.8?
k=log4 91.8 4^k=91.8 -- b/c of log rules-- log 4^k=log 91.8 -- b/c of log rules-- k*log 4=log91.8 --> divide by log 4 k=log 91.8/log 4 k= 3.260
Log 4 64 equals y?
Log4 64=y 64=4y 26=22y Therefore y=3
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.
A logarithmic function is the same as an exponential function?
Apex: false A logarithmic function is not the same as an exponential function, but they are closely related. Logarithmic functions are the inverses of their respective exponential functions. For the function y=ln(x), its inverse is x=ey For the function y=log3(x), its inverse is x=3y For the function y=4x, its inverse is x=log4(y) For the function y=ln(x-2), its inverse is x=ey+2 By using the properties of logarithms, especially the fact that a number raised to a logarithm of base itself equals the argument of the logarithm: aloga(b)=b you can see that an exponential function with x as the independent variable of the form y=f(x) can be transformed into a function with y as the independent variable, x=f(y), by making it a logarithmic function. For a generalization: y=ax transforms to x=loga(y) and vice-versa Graphically, the logarithmic function is the corresponding exponential function reflected by the line y = x.
Can anyone solve for x log 4 X plus 16 plus log 4 X plus 4 equals 3?
log4(x) +16 +log4(x) +4=32log4(x)=-17log4(x)=-17/2x=4^(-17/2)=========================Since the parentheses have been lost from the question,it could easily be interpreted this way instead, (as well asa few others):x log(4x) + 16 + log(4x) + 4 = 3(x + 1) log(4x) = -174x = 10-17/(x+1)4x = the (x+1)th root of 10-17Come on back and solve that one for us.
