1. Aug 6, 2010

### theanimux

1. The problem statement, all variables and given/known data

Solve for x

(1/10)^(x - 1) < (1/10)

2. My Dilemma

The problem is that using one method will yield one answer, but using another method will yield another. Why is that? Please help.

My issue is not regarding the right answer (the right answer is x > 2). It's regarding the method/mathematical operation. Thanks!

3. Solution attempts

Method 1: Multiplying exponents (this method should be the correct one)
Step 1: (10^-1)^(x - 1) < 10^-1
Step 2: multiply exponents:: 10^(-x + 1) < 10^-1
Step 3: simplify:: -x + 1 < -1
Step 4: solve:: x > 2

Method 2: Please tell me what's wrong with this method
Step 1: [(1/10)^x] / (1/10) < (1/10)
Step 2: multiply both sides by (1/10):: (1/10)^x < (1/10)^2
Step 3: solve:: x < 2

2. Aug 6, 2010

### N-Gin

Step 3 is wrong. From the step 2, you actually conclude that x > 2. If the base is less than zero, the inequality sign "changes direction" (as it does when you multiply inequality with -1).

3. Aug 6, 2010

### theanimux

Do you mean when the base < 1? I tested cases, and when the base < 1, I would need to reverse the sign.

And when the base > 1, I can just continue order of operations like any normal equation.

But when base = 1, then it appears that there is no solution. Do you have an explanation for that as well?

Thanks a lot!

4. Aug 6, 2010

### Dick

You are trying to solve an equation like a^x<a^2. You are doing it by implicitly taking logs. So x*log(a)<2*log(a). Now you want to divide by log(a). If log(a) is positive (a>1) you get x<2. If log(a) is negative (0<a<1) you have to reverse the inequality, so you get x>2. If a=1 then the original inequality is 1^x<1^2. To think about that problem, you don't need logs or anything.

5. Aug 7, 2010

### Anakin_k

.1^(x-1) < .1
log (.1)^(x-1) < log .1
(x-1)(log .1) < log .1

Expand and solve.

6. Aug 8, 2010

### Mentallic

Logs aren't necessary.

$$\left(\frac{1}{10}\right)^x<\left(\frac{1}{10}\right)^2$$

$$\frac{1}{10^x}<\frac{1}{10^2}$$

Multiply through:

$$10^2<10^x$$

$$x>2$$ as required.

Or you can even use the property that $$\frac{1}{a^x}=a^{-x}$$