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

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).

Recognize that -0,04 is $-\frac{1}{25} = -\frac{1}{5^2}$ – Malcolm Mar 17 '18 at 18:19 — Malcolm, Mar 17 '18 at 18:19
@Malcolm Why is it equal to $-\frac{1}{25}$? – Mar 17 '18 at 18:20 — , Mar 17 '18 at 18:20
$-0,04 = -\frac{4}{100} = -\frac{4}{4\cdot 25}$ – Malcolm Mar 17 '18 at 18:21 — Malcolm, Mar 17 '18 at 18:21

score 1 · Answer 1 · answered Mar 17 '18 at 18:20

1

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

answered Mar 17 '18 at 18:20

Dr. Sonnhard Graubner

95,283

score 1 · Answer 2 · answered Mar 17 '18 at 18:20

1

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

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

answered Mar 17 '18 at 18:20

lab bhattacharjee

274,582

I still didn't get it. – Mar 17 '18 at 18:22
@Busi, Where is ur confusion? – lab bhattacharjee Mar 17 '18 at 18:23
At the first step you've done. – Mar 17 '18 at 18:23
@Busi, $\dfrac ab-\dfrac cd=?$ – lab bhattacharjee Mar 17 '18 at 18:25
We're finding a common denominator here. $$\frac 4{5^{x+1}} - \frac 1{5^x} = \frac {4}{5^{x+1}} - \frac 5{5\cdot 5^x} = \frac{4}{5^{x+1}}-\frac 5{5^{x+1}} = \cdots$$ – amWhy Mar 17 '18 at 18:25

score 1 · Answer 3 · answered Mar 17 '18 at 18:20

1

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

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

$$x+1=2$$

$$x=1$$

answered Mar 17 '18 at 18:20

CY Aries

23,393

score 1 · Answer 4 · answered Mar 17 '18 at 19:24

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

Check:

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

Dylan Zammit · Answer 5 · 2018-03-17T18:27:16.080

0

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

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

edited Mar 17 '18 at 18:27

answered Mar 17 '18 at 18:21

Dylan Zammit

1,218

How did you get $4-5=-0.04 \cdot 5^{x+1}$? – Mar 17 '18 at 18:23
multiplied both sides by $5^{x+1}$, the terms cancel – Dylan Zammit Mar 17 '18 at 18:23
Can you show that? – Mar 17 '18 at 18:24
updated the solution – Dylan Zammit Mar 17 '18 at 18:25
Now why isn't it $-\frac{1.5^{x+1}}{5^x}$? – Mar 17 '18 at 18:26
because i made a mistake :) nice catch. updated again – Dylan Zammit Mar 17 '18 at 18:27
Oh, got it perfectly! However, the last thing I didn't get is $4-5=-0.04 \cdot 5^{x+1}$ If you can explain this one too, I'd be grateful. – Mar 17 '18 at 18:28
1

@Busi Note that $5^{x+1}=5\cdot 5^x$ – Mar 17 '18 at 18:31

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

5 Answers5