$\frac{4}{5^{x+1}}-\frac{1}{5^x} = -0.04$

Question

$$\frac{4}{5^{x+1}}-\frac{1}{5^x} = -0.04$$

This equality seems too simple to solve but I've to know what to do.

The first thing I thought is $\frac{1}{5^x} = 5^{-x}$.

However, no idea about others.

Instead of solving this question, I'd like to get your tips/hints for the terms like $-0.04$ (which are making me very very confused).

Answer 1

write $$\frac{4}{5\cdot 5^x}-\frac{1}{5^x}=-\frac{4}{100}$$ and Substitute $$5^x=t$$ or $$\frac{4}{5\cdot 5^x}-\frac{5}{5\cdot 5^x}=-\frac{4}{100}$$

Answer 2

$$\dfrac4{5^{x+1}}-\dfrac1{5^x}=\dfrac{4-5}{5^{x+1}}$$

Now, $0.04=\dfrac4{100}=?$

Answer 3

$$\left(4-5\right)\frac{1}{5^{x+1}}=-0.04$$

$$-\frac{1}{5^{x+1}}=-\frac{1}{25}$$

$$x+1=2$$

$$x=1$$

Answer 4

Multiply both sides by $5^{x+1}$ and you get:

$$\dfrac{4\cdot 5^{x+1}}{5^{x+1}}-\dfrac{1\cdot 5^{x+1}}{5^{x}}=-0.04 \cdot 5^{x+1}$$ $$4-5=-0.04 \cdot 5^{x+1}$$ $$\implies 25=5^{x+1}$$ $$\implies x=1$$

Answer 5

We have $$\begin{align}\frac{4}{5^{x+1}}-\frac{1}{5^x} = -0.04&\implies5^{x+1}\left(\frac{4}{5^{x+1}}-\frac{1}{5^x}\right)=4-5=-0.04\cdot5^{x+1}\\&\implies5^{x+1}=\frac1{0.04}=25=5^2\\&\implies x+1=2\\&\implies x=1\end{align}$$

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Check:

$$\frac4{5^{1+1}}-\frac1{5^1}=\frac4{25}-\frac5{25}=-\frac1{25}=-0.04$$