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Rules of log?
Here are a few, note x>0 and y>0 and a&b not = 1 * log (xy) = log(x) + log(y) * log(x/y) = log(x) - log(y) * loga(x) = logb(x)*loga(b) * logb(bn) = n * log(xa) = a*log(x) * logb(b) = 1 * logb(1) = 0
How do I solve the equation e to the power of x equals 2 using logarithms?
When the logarithm is taken of any number to a power the result is that power times the log of the number; so taking logs of both sides gives: e^x = 2 → log(e^x) = log 2 → x log e = log 2 Dividing both sides by log e gives: x = (log 2)/(log e) The value of the logarithm of the base when taken to that base is 1. The logarithms can be taken to any base you like, however, if the base is e (natural logs, written as ln), then ln e = 1 which gives x = (ln 2)/1 = ln 2 This is in fact the definition of a logarithm: the logarithm to a specific base of a number is the power of the base which equals that number. In this case ln 2 is the number x such that e^x = 2. ---------------------------------------------------- This also means that you can calculate logs to any base if you can find logs to a specific base: log (b^x) = y → x log b = log y → x = (log y)/(log b) In other words, the log of a number to a given base, is the log of that number using any [second] base you like divided by the log of the base to the same [second] base. eg log₂ 8 = ln 8 / ln 2 = 2.7094... / 0.6931... = 3 since log₂ 8 = 3 it means 2³ = 8 (which is true).
What is the x and y equal if x - 3y equals 3?
There are an infinite number of them. Here are a few: x= -2, y= -5/3 x= -1, y= -4/3 x= 0, y= -1 x= 1, y= -2/3 x= 2, y= -1/3 x= 3, y= 0 x= 12, y= 3
What is x when log base 3 x - 2log base x 3 equals 1?
log3x - 2logx3 = 1 so 1/logx3 - 2logx3 = 1 Let y = logx3 then 1/y - 2y = 1 multiply through by y: 1 - 2y2 = y that is, 2y2 + y - 1 = 0 which factorises to (y + 1)(2y -1) = 0 so that y = -1 or y = 1/2 y = -1 implies logx3 = -1 so that x-1 = 3 ie 1/x = 3 or x= 1/3 y = 1/2 implies logx3 = 1/2 so that x1/2 = 3 or x = 32 = 9
2x equals 3-y 2y equals 12-x?
2x=3-y 2y=12-x y=3-2x 2(3-2x)=12-x 6-4x=12-x -3x=6 x=-2 2(-2)=3-y -4=3-y y=7 x=-2 y=7
Give an example of an exponential function Find this exponential function's inverse which will be a logarithmic function Plot the graph of both the functions?
An exponential function can be is of the form f(x) = a*(b^x). Some examples are f1(x) = 3*(10^x), or f2(x) = e^(-2*x). Note that the latter still fits the format, with b = e^(-2). The inverse is the logarithmic function. So for y = f1(x) = 3*(10^x), reverse the x & y, and solve for y:x = 3*(10^y)log(x) = log(3*(10^y)) = log(3) + log(10^y) = log(3) + y*log(10) = y*1 + log(3)y = log(x) - log(3) = log(x/3)The second function: y = e^(-2*x), the inverse is: x = e^(-2*y).ln(x) = ln(e^(-2*y)) = -2*y*ln(e) = -2*y*1y = -ln(x)/2 = ln(x^(-1/2))See related link for an example graph.
How do you find cube of a number using log?
If log(x) = y then log(x3) = 3*log(x) = 3*y so that x3 = antilog(3*y) So, to find the cibe of x 1) find log x 2) multiply it by 3 3) take the antilog of the result.
What are 3 logarithmic properties?
log(1) = 0log(x*y) = log(x) + log(y)If logb(x) = y then x = by.
How do you half a number by using only powers?
You can't: let suppose y the power of x to obtain such a result then xy=x/2 then xy-1=1/2 (y-1) log (x) = - log(2) (if x is a positive number) y-1 = -log(2)/log(x) y = 1 - log(2)/log(x) So log function must also being used!
Why can't the base on logarithms be one?
let's look at log base 2 (x)=8, that means 2^x=8 so x=3 in general if b^y=x, then log base b (x)=y if the base is 1, then we have 1^y=x, but 1^y=1 for all y so it does not work..
What is the inverse of y equal the log of 2x?
y = log 2x → x = 1/2 <base of log>y So: y = log102x → x = 1/210y (common logs) y = loge2x → x = 1/2ey (natural logs)
If log y equals 0.7 plus 2 times log x express y in terms of x?
It cannot be done because the base for the second log is not given.
What is inverse of y equals 2radical x plus 3?
To find the inverse you switch the x and the y and then solve for y. x=2 radical( y + 3) radical(y + 3) = x/2 y+3= (x/2)² y = (x/2)² -3 So the answer is y = (x/2)² -3
If y equals 10 then what is then what is y equals log x?
y = 10 y = log x (the base of the log is 10, common logarithm) 10 = log x so that, 10^10 = x 10,000,000,000 = x
Rules of log?
Here are a few, note x>0 and y>0 and a&b not = 1 * log (xy) = log(x) + log(y) * log(x/y) = log(x) - log(y) * loga(x) = logb(x)*loga(b) * logb(bn) = n * log(xa) = a*log(x) * logb(b) = 1 * logb(1) = 0
The log of a quotient is the log of a numerator divided by the log of the denominator?
For a quotient x/y , then its log is logx - log y . NOT log(x/y)
From the equation X-Y equals 2 what is Y when X equals 3?
X - Y = 2 When X = 3 you have 3 - Y = 2 Moving Y to RHS: 3 = Y + 2 Moving 2 to LHS: 1 = Y
