Find the function of the form y = log(base a) x whose graph is given?
Using the point you are given, (3, 1/2) then log(base a) (3) = 1/2 Writing as an exponential a^(1/2) =3 [a^(1/2)]^2 = 3^2 a = 3^2 = 9
Find the function of the form y = log(base a) x whose graph is given. Please help me?
Hint: c=3 , and it's easy to do the rest
Find the function of the form y = loga x whose graph is given.?
It's y= log4x, with 4 being the base, because the equation becomes a^y=x, and when 4^-1, x=1/4, which is the point you gave.
Find the function of the form y = log a (x) whose graph is given (6,1)?
1 = log_a(6) a = 6 so y = log_6(x) ♣♦
Find the function of the form y=log base of a to the x whose graph is given?
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How do I find a function of y = loga x?
first question: all logarithmic functions have a vertical aysmptote x=0 and passes through (1,0) need the point (9,2) plug in the eqn: 2= loga 9 a^2= 9 {write log in exponential form] a=3 so the function is y=log3 x The other problem you stated is a polynomial function.. same idea on solving for a b and c the approach will be different in terms of solving for these unknown. since when x and y are plugged in, you'll be solving for linear equations instead of exponential equations just like the problem above
Write a power function of the form y=ax^b whose graph passes through the given point. (4, 7) (2, 5)?
7 = a*4^b 5 = a*2^b 7/4^b = a (5/7)^(1/b) = (1/2) (5/7) = (1/2)^b ln(5/7) = b*ln(1/2) b = ln(5/7)/ln(1/2) b = 0.4854 a = 7/4^(0.4854) = 3.571 y = 3.571*x^(0.4854) The answer is very close, but these values are rounded, so if you plug in the values 4 and 2 for x, you should get values very close to 7 and 5, but not quite due to rounding. If you want to exact answer, leave the constants as they are without turning them into decimals. There you go. Hope that helps.
How do I find the base of this log function?
logarithmic: log [base x] (3) = 0.5 Exponential: x^(0.5) = 3 or √x = 3 x = 9 rather easy...really !!
Find the base of exponential function whose graph contains the given points?
Well, I don't know if you can see it straight away, but, you can work it out by first writing the function in the desired form, with unknown constants. i.e. y = a^x where a is some constant. Now we substitute in (x,y) = (-2 ,1/16): 1/16 = a^-2 1/16 = 1/a^2 16 = a^2 a = +4 or -4 So the function could be: y = 4^x or y = (-4)^x Usually we don't consider negative values of a to be exponential functions, so the answer you were probably looking for is y = 4^x
Note that this is the same as asking whether there exists a solution to the differential equation [math] y’ = y [/math], given some initial condition [math] y(0) = y_0 [/math] (if you want to recover [math] e^x [/math] this way, rather than, say, [math] 0 [/math], you would want to specify that [math] y(0) = 1 [/math]).In general, proving that a differential equation must have a solution is incredibly difficult. But this isn’t just any differential equation—this is a first order differential equation. That is, it is something of the form [math] y’ = f(x,y) [/math]. It turns out that if [math] f(x,y) [/math] is a smooth function, then such an equation always has a smooth solution for any initial condition.Intuitively, this isn’t so hard to see. Draw the slope field for this differential equation. The solution to this equation will be a curve that is tangent to the slope field at every point. Start at your initial condition [math] (0, y_0) [/math]—since [math] f [/math] is smooth, if you look at a very small patch around this point, the slope field will look essentially straight, and so it is easy to see that there must be a curve that is tangent to it at every point. So draw a little bit of this curve, and then look at a small patch at where you left off. Repeat. You have constructed the desired solution.To turn this into a fully rigorous proof, you can use the Inverse function theorem to turn the slope field into a straight field in small neighborhoods—draw the solution there, and then invert back.