Solve 0,2^x+0,5*25/sqrt{5}=0,04^x

$$\frac { 0,2 ^ { x + 0,5 } \cdot 25 } { \sqrt { 5 } } = 0,04 ^ { x }$$

Solve for x

$$x\in \mathrm{R}$$

Solution Steps

Multiply 0 and 25 to get 0.

$$\frac{0\times 2^{x+0,5}}{\sqrt{5}}=0\times 4^{x}$$

Rationalize the denominator of \frac{0\times 2^{x+0,5}}{\sqrt{5}} by multiplying numerator and denominator by \sqrt{5}.

$$\frac{0\times 2^{x+0,5}\sqrt{5}}{\left(\sqrt{5}\right)^{2}}=0\times 4^{x}$$

The square of \sqrt{5} is 5.

$$\frac{0\times 2^{x+0,5}\sqrt{5}}{5}=0\times 4^{x}$$

Divide 0\times 2^{x+0,5}\sqrt{5} by 5 to get 0\times 2^{x+0,5}\sqrt{5}.

$$0\times 2^{x+0,5}\sqrt{5}=0\times 4^{x}$$

Subtract 0\times 4^{x} from both sides.

$$0\times 2^{x+0,5}\sqrt{5}-0\times 4^{x}=0$$

Reorder the terms.

$$\text{true}$$

This is true for any x.

$$x\in \mathrm{R}$$