Solve 0=(x-t)*(e^0,2x-1)

$$0 = ( x - t ) \cdot ( e ^ { 0,2 x } - 1 )$$

Solve for t

$$\left\{\begin{matrix}\\t=x\text{, }&\text{unconditionally}\\t\in \mathrm{R}\text{, }&x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply x-t by e^{0,2x}-1.

$$0=xe^{0,2x}-x-te^{0,2x}+t$$

Swap sides so that all variable terms are on the left hand side.

$$xe^{0,2x}-x-te^{0,2x}+t=0$$

Subtract xe^{0,2x} from both sides. Anything subtracted from zero gives its negation.

$$-x-te^{0,2x}+t=-xe^{0,2x}$$

Add x to both sides.

$$-te^{0,2x}+t=-xe^{0,2x}+x$$

Combine all terms containing t.

$$\left(-e^{0,2x}+1\right)t=-xe^{0,2x}+x$$

The equation is in standard form.

$$\left(1-e^{\frac{x}{5}}\right)t=x-xe^{\frac{x}{5}}$$

Divide both sides by -e^{0,2x}+1.

$$\frac{\left(1-e^{\frac{x}{5}}\right)t}{1-e^{\frac{x}{5}}}=\frac{x-xe^{\frac{x}{5}}}{1-e^{\frac{x}{5}}}$$

Dividing by -e^{0,2x}+1 undoes the multiplication by -e^{0,2x}+1.

$$t=\frac{x-xe^{\frac{x}{5}}}{1-e^{\frac{x}{5}}}$$

Divide -xe^{\frac{x}{5}}+x by -e^{0,2x}+1.

$$t=x$$

Solve for x

$$x=0$$

$$x=t$$