Q: If f x log 15 x what is f-1 x?

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log(x) - log(6) = log(15)Add log(6) to each side:log(x) = log(15) + log(6) = log(15 times 6)x = 15 times 6x = 90

f(x)=ln3x f1(x)=3/3x=1/x

If f(x)=1/x then F(x)=antiderivative of f(x)=ln(|x|) (the natural log of the absolute value of x) There's another way of reading this question. The anti derivative of 1 is x+c. Dividing that by x gives you 1 + c/x

log(x) + log(2) = log(2)Subtract log(2) from each side:log(x) = 0x = 100 = 1

Not quite. The log(x/y) = log(x) - log(y) In words, this reads "The log of a quotient is the difference of the log of the numerator and the log of the denominator."

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log(x) - log(6) = log(15)Add log(6) to each side:log(x) = log(15) + log(6) = log(15 times 6)x = 15 times 6x = 90

2*log(15) = log(x) 152 = x; its equivalent logarithmic form is 2 = log15 x (exponents are logarithms) then, it is equivalent to 2log 15 = log x, equivalent to log 152 = log x (the power rule), ... 2 = log15 x 2 = log x/log 15 (using the change-base property) 2log 15 = log x Thus, we can say that 152 = x is equivalent to 2*log(15) = log(x) (equivalents to equivalents are equivalent)

try this out...#include#includefloat valcal(float);int main(){float a,b,f,x,f1,f2,f3,i;a=2;b=3;int flag=0;f1=valcal(a);f2=valcal(b);printf("\n\n%f %f",f1,f2);for(i=0;i

f(x)=ln3x f1(x)=3/3x=1/x

try this out...#include#includefloat valcal(float);int main(){float a,b,f,x,f1,f2,f3,i;a=2;b=3;int flag=0;f1=valcal(a);f2=valcal(b);//printf("\n\n%f %f",f1,f2);for(i=0;i

To find the acceleration, first resolve the forces F1 and F2 into their x and y components. The x-component of F1 is 20 N and the x-component of F2 is 15 N * cos(60 deg). The y-component of F2 is 15 N * sin(60 deg). The net force in the x-direction is 20 N + 7.5 N, and the net force in the y-direction is 15 N * sin(60 deg). Then, calculate the net force by using the Pythagorean theorem on the x and y components. Finally, divide the net force by the mass of 8 kg to get the acceleration.

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.

I'm not sure I really understand the question. If you have two functions, F1 and F2, then by definition of function you have: If X=Y then F(x)=F(y) for both F1 and F2 (F1+F2)(x) is defined as F1(X) + F2(X), and for every X there exists a unique F1(x) and F(2)(x) and the sum of these two numbers is obviously unique. If the question meant that the range of F1 and F2 were included in Z and how do you show that the range of F1 + F2 is also a subset of Z, then it's as simple as: the sum of two integers is an integer. and for the second part: the product of two integers is an integer.

PSMF(x) %% Implementation of Progressive Switching Median Filter %% Base Paper : Zhou Wang and David Zhang, "Progressive Switching Median %% Filter for the Removal of Impulse Noise from Highly Corrupted Images", %% IEEE Trans. on Cir. and Sys., vol. 46, no. 1, Jan. 1999. %% Function Y = PSMF(x) %% input x = Image is corrupted by Salt & Pepper Noise %% %% Example: Y = PSMF(x); %% Posted date : 16 - 10 - 2008 %% Modified date : %% %% Developed By : K.Kannan (kannan.keizer@gmail.com) %% & Jeny Rajan (jenyrajan@gmail.com) %% Medical Imaging Research Group (MIRG), NeST, %% Trivandrum. %% Progressive Switching Median Filter function Y = PSMF(x) x = double(x); WF = 3; ND = 3;T = 40;a = 65;b = -50; M = medfilt2(x,[3 3]); N = abs(x - M); N(N>T)=0; N = N ~= 0; N = double(N); R = sum(N(:))/(size(x,1) * size(x,2)); if R <= 0.25 WD = 3; else WD = 5; end TD = a + (b * R); z = IMPDET(x,ND,WD,TD); Y = NF(x,z,WF); %% Impulse Detection function F1 = IMPDET(x,ND,WD,TD) X = x; M = medfilt2(X,[WD WD]); D = abs(X - M); F = zeros(size(x)); F(D>=TD)=1; F1 = F; X(F1==F)=X(F1==F); X(F1~=F)=M(F1~=F); for i = 1:ND-1 M = medfilt2(X,[WD WD]); F1(abs(X - M)<TD)=F(abs(X - M)<TD); F1((X - M)>=TD)=1; X(F1==F)=X(F1==F); X(F1~=F)=M(F1~=F); F = F1; end return; %% Noise Filtering function Y = NF(x,f,WF) g = f; Y = x; Y1 = Y; g1 = g; s = sum(g(:)); while s ~= 0 M = medfilt2(Y,[WF WF]); Y1(g==1)=M(g==1); g1(Y~=Y1)=0; Y = Y1; g = g1; s1 = sum(g(:)); if s1 ~= s s = s1; else s = 0; end end return;

The only way I ever learned to find it was to think about it. The function f(x) = log(x) only exists of 'x' is positive. As 'x' gets smaller and smaller, the function asymptotically approaches the y-axis.

It has no specific name. For example f(x) = sin(x)/log(x) where x not equal to 1

#include#include #include#define ESP 0.001#define F(x) (x)*(x)*(x) + (x)*(x) + (x) + 7void main(){int i=1;float x0,x1,x2;double f1,f2,f0,t;clrscr( );printf( "\nEnter the value of x0: ");scanf( "%f",&x0);printf( "\nEnter the value of x1: ");scanf( "%f",&x1);printf( "\n____________________________________________\n");printf( "\niteration\tx0\tx1\tx2\tf0\tf1\tf2");printf( "\n_____________________________________________\n");do{x2= (x0+x1)/2;f0= F(x0);f1= F(x1);f2= F(x2);printf( "\n%d %f %f %f %lf %lf %lf",i,x0,x1,x2,f0,f1,f2);if (f0*f2ESP);printf( "\n________________________________________\n");printf( "\n\nApp.root = %f",x2);getch( );}