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I am having difficult time with a contradiction. Here is the simple math problem that I cannot understand why exactly the same technique gives two different (and contradictory) results:

The question is what is the smallest x value that satisfies the below inequality:

$$(\frac{2}{3})^{2x-1}< (\frac{27}{8})^{x-2}$$

if you rearrange the left item and solve the problem as:

$$(\frac{3}{2})^{-2x+1}<(\frac{3}{2})^{3x-6}$$ $$-2x+1<3x-6$$ $$7<5x$$ $$\frac{7}{5} < x$$ which yields 2 as the answer. However, if you rearrange the right item of the inequality the steps are:

$$(\frac{2}{3})^{2x-1}<(\frac{2}{3})^{6-3x}$$ $$2x-1<6-3x$$ $$5x<7$$ $$\frac{7}{5} > x$$ which yields 1 as the answer. where is the problem? What is the cause of this contradiction? Thank you.

UKadir
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    You have in effect found the only solution of the equality is when $x=\frac75$. When $x$ is large and positive, the inequality is satisfied as the left hand side is less than $1$ and the right hand side greater than $1$; for similar reasons the inequality is not satisfied when $x$ is large and negative. So you first approach gets the correct answer. Your second has an error of ignoring that $\frac23 <1$ and $\log(2/3) <0$ so reversing the inequality. You should have $\left(\frac{2}{3}\right)^{2x-1}<\left(\frac{2}{3}\right)^{6-3x} \implies 2x-1>6-3x$ – Henry May 11 '21 at 11:06
  • Aside: your jump from $\frac75$ to $2$ or $1$ is implying that $x$ must be an integer, which you've not otherwise said. On the other hand, "the smallest $x$ such that $x>\frac75$" doesn't really make sense, because there is no such $x$ (instead of "smallest", you could ask for the "greatest lower bound" of all such $x$ values). (Also, the smallest $x$ such that $x<\frac75$ is negative infinity, by that same interpretation.) – Teepeemm May 11 '21 at 19:20

2 Answers2

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Note that $x\to b^x$ is strictly increasing when $b>1$ and therefore $$b^x<b^y\Leftrightarrow x<y.$$ You applied this property correctly for $b=3/2>1$.

On the other hand, $x\to b^x$ is strictly decreasing when $0<b<1$ and therefore we have $$b^x<b^y\Leftrightarrow x>y.$$ Hence for $b=2/3<1$ you should find the reversed inequality $$\left(\frac{2}{3}\right)^{2x-1}<\left(\frac{2}{3}\right)^{6-3x}\Leftrightarrow 2x-1>6-3x$$ which leads to the same result.

Robert Z
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Yes, you have to invert the sign of inequality while taking logarithm depending upon whether base of log $>1$ or $<1$. This is because $a^x$ is an increasing function if $a>1$, but it is a decreasing function if $a<1$, hence log with respect to $a$ follows a similar behavior. If you invert the inequality sign while taking log with respect to base $\frac 23$, answer would match.

Ritam_Dasgupta
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