Solve (x+hx)-f(x)/hf(x)=e^frac{1{x}}

$$\frac { ( x + h x ) - f ( x ) } { h f ( x ) } = e ^ { \frac { 1 } { x } }$$

Solve for f

$$f=\frac{h+1}{he^{\frac{1}{x}}+1}$$

$$x\neq 0\text{ and }h\neq -1\text{ and }\left(h>0\text{ or }x\neq \frac{1}{\ln(-\frac{1}{h})}\right)\text{ and }h\neq 0$$

Steps for Solving Linear Equation

Variable f cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by fhx.

$$x+hx-fx=fhxe^{\frac{1}{x}}$$

Subtract fhxe^{\frac{1}{x}} from both sides.

$$x+hx-fx-fhxe^{\frac{1}{x}}=0$$

Subtract x from both sides. Anything subtracted from zero gives its negation.

$$hx-fx-fhxe^{\frac{1}{x}}=-x$$

Subtract hx from both sides.

$$-fx-fhxe^{\frac{1}{x}}=-x-hx$$

Combine all terms containing f.

$$\left(-x-hxe^{\frac{1}{x}}\right)f=-x-hx$$

The equation is in standard form.

$$\left(-hxe^{\frac{1}{x}}-x\right)f=-hx-x$$

Divide both sides by -x-hxe^{x^{-1}}.

$$\frac{\left(-hxe^{\frac{1}{x}}-x\right)f}{-hxe^{\frac{1}{x}}-x}=-\frac{x\left(h+1\right)}{-hxe^{\frac{1}{x}}-x}$$

Dividing by -x-hxe^{x^{-1}} undoes the multiplication by -x-hxe^{x^{-1}}.

$$f=-\frac{x\left(h+1\right)}{-hxe^{\frac{1}{x}}-x}$$

Divide -x\left(1+h\right) by -x-hxe^{x^{-1}}.

$$f=\frac{h+1}{he^{\frac{1}{x}}+1}$$

Variable f cannot be equal to 0.

$$f=\frac{h+1}{he^{\frac{1}{x}}+1}\text{, }f\neq 0$$

Solve for h

$$h=-\frac{1-f}{1-fe^{\frac{1}{x}}}$$

$$x\neq 0\text{ and }f\neq 1\text{ and }\left(f<0\text{ or }x\neq -\frac{1}{\ln(f)}\right)\text{ and }f\neq 0$$

Steps for Solving Linear Equation

Variable h cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by fhx.

$$x+hx-fx=fhxe^{\frac{1}{x}}$$

Subtract fhxe^{\frac{1}{x}} from both sides.

$$x+hx-fx-fhxe^{\frac{1}{x}}=0$$

Subtract x from both sides. Anything subtracted from zero gives its negation.

$$hx-fx-fhxe^{\frac{1}{x}}=-x$$

Add fx to both sides.

$$hx-fhxe^{\frac{1}{x}}=-x+fx$$

Combine all terms containing h.

$$\left(x-fxe^{\frac{1}{x}}\right)h=-x+fx$$

The equation is in standard form.

$$\left(x-fxe^{\frac{1}{x}}\right)h=fx-x$$

Divide both sides by x-fxe^{x^{-1}}.

$$\frac{\left(x-fxe^{\frac{1}{x}}\right)h}{x-fxe^{\frac{1}{x}}}=\frac{x\left(f-1\right)}{x-fxe^{\frac{1}{x}}}$$

Dividing by x-fxe^{x^{-1}} undoes the multiplication by x-fxe^{x^{-1}}.

$$h=\frac{x\left(f-1\right)}{x-fxe^{\frac{1}{x}}}$$

Divide x\left(-1+f\right) by x-fxe^{x^{-1}}.

$$h=\frac{f-1}{1-fe^{\frac{1}{x}}}$$

Variable h cannot be equal to 0.

$$h=\frac{f-1}{1-fe^{\frac{1}{x}}}\text{, }h\neq 0$$