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What is the expression with the logarithm Log 4 plus Log 3?
log4+log3=log(4x3)=log12
The value of log log4log4x equals 1 then x equals?
the value of log (log4(log4x)))=1 then x=
What is the value of x that satisfies the equation 2 log 4 Bracket 7 x Bracket - 5 equals 0 Is - Please show working out?
2log4(7x) - 5 = 0 2log4(7x) = 5 log4(7x) = 5/2 7x = 45/2 =(41/2)5 = 25 = 32 x = 32/7
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.
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
How do you express log4 7 as a single logarithm with base 2?
The rule for converting log bases is: log4(7) = log2(7)/log2(4) Now 4 = 2 squared so log2(4) = 2 So log4(7) = log4(7)/2 = (1/2)*log2(7) By the way, the "new and improved" browser now does not accept subscripts or superscripts so I hope you understand this.
What is log4 19 to four right decimal places?
It is 2.1240
The value of log log4log4x equals 1 then x equals?
the value of log (log4(log4x)))=1 then x=
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
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.
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 x if the question is log4 7x5?
To solve for ( x ) in the equation ( \log_4 7x = 5 ), we can rewrite it in exponential form: ( 7x = 4^5 ). Calculating ( 4^5 ) gives us ( 1024 ), so we have ( 7x = 1024 ). Dividing both sides by 7 results in ( x = \frac{1024}{7} ), which simplifies to approximately ( 146.29 ).
What is the value of x that satisfies the equation 2 log 4 Bracket 7 x Bracket - 5 equals 0 Is - Please show working out?
2log4(7x) - 5 = 0 2log4(7x) = 5 log4(7x) = 5/2 7x = 45/2 =(41/2)5 = 25 = 32 x = 32/7
What kind of sequence is formed by log2 log4 log8?
The sequence formed by ( \log_2 4 ) and ( \log_2 8 ) can be evaluated as follows: ( \log_2 4 = 2 ) (since ( 4 = 2^2 )) and ( \log_2 8 = 3 ) (since ( 8 = 2^3 )). Therefore, the sequence is ( \log_2 2 = 1, \log_2 4 = 2, \log_2 8 = 3 ). This forms an arithmetic sequence with a common difference of 1.
