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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.
