Confusing Problem - Equation w/ Negative Exponent

[theanimux]

The Problem: Solve for x

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

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

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

 

[soroban]

Hello, theanimux!

\(\displaystyle \text{Solve for }x\!:\;\;\left(\frac{1}{10}\right)^{x - 1} \;<\; \frac{1}{10}\)


\(\displaystyle \text{Method 1: Multiplying exponents}\)

. . \(\displaystyle \begin{array}{ccccc}\text{We have:} & (10^{-1})^{x - 1} &<& 10^{-1} \\ \\[-3mm] \text{Multiply exponents:} & 10^{-x + 1} &<& 10^{-1} \\ \text{Equate exponents;} & -x + 1 &<& -1 \\ \text{Solve:} & x &>& 2 \end{array}\)


\(\displaystyle \text{Method 2: what's wrong with this method?}\)

. . \(\displaystyle \begin{array}{ccccc}\text{We have:} & \left(\frac{1}{10}\right)^x\left(\frac{1}{10}\right)^{-1} & < & \frac{1}{10} \\ \\[-3mm] \text{Multiply by }\frac{1}{10}: & \left(\frac{1}{10}\right)^x &<& \left(\frac{1}{10}\right)^2 \\ \\[-3mm] \text{Solve:} & x &<& 2 \end{array}\)

There is a very subtle error . . . It can be blamed on our teachers.
. . [I was a math professor for 36 years . . . I accept some of the blame.]


\(\displaystyle \text{We are told: }\;\text{If }b^x \,=\,b^y,\,\text{ we can equate the exponents: }\;x\,=\,y\)


\(\displaystyle \text{But we were }\bf{not}\text{ warned about this:}\)

. . \(\displaystyle \text{If }b^x \,<\:b^y,\:\text{ then }x\,<\,y\;\;\hdots\;\; \bf{provided\;that\;\;b \,>\,1}\)

. . \(\displaystyle \text{If } b \,<\,1,\;\text{ the inequality must be reversed.}\)

 

[theanimux]

Thank you so much : )

That clarifies for cases b > 1 and b < 1, but what about for when b = 1?

I've tested cases for when
1^(x-1)<1
:: x-1 < 1
:: x<2

but in this situation, 1^(x-1) is never < 1, regardless of what value x is. Could you explain the rule for when base = 1?

Thank you so much!!!

 

[mmm4444bot]

Thank you so much : )

what about when b = 1?

All powers of 1 are equal, so b^x can never be less than b^y, when b = 1.

In other words, the rule posted by Soroban does not apply, when b = 1. It only applies to situations where b is not 1.

 

[mmm4444bot]

\(\displaystyle \text{Method 2: what's wrong with this method?}\)

. . \(\displaystyle \begin{array}{ccccc}\text{We have:} & \left(\frac{1}{10}\right)^x\left(\frac{1}{10}\right)^{-1} & < & \frac{1}{10} \\ \\[-3mm] \text{Multiply by }\frac{1}{10}: & \left(\frac{1}{10}\right)^x &<& \left(\frac{1}{10}\right)^2 \\ \\[-3mm] \text{Solve:} & x &<& 2 \end{array}\)

Umm, you skipped some steps? :wink:

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

\(\displaystyle (10^{-1})^x \;<\; (10^{-1})^2\)

\(\displaystyle 10^{-x} \;<\; 10^{-2}\)

\(\displaystyle -x \;<\; -2\)

\(\displaystyle x \;>\; 2\)