# Thread: Which direction for exponential inequality?

1. ## Which direction for exponential inequality?

If
$\ln(x) < y$,

then fill in the blank (which direction does the inequality go)
$x \qquad e^{y}$

(And please explain how you got that).

2. ## Re: Which direction for exponential inequality?

You are posting in a precalc forum, so I am not sure exactly what information you have to work with. Do you know that the function $e^x$ is a strictly increasing function? In other words, if $x then $e^x?

3. ## Re: Which direction for exponential inequality?

Originally Posted by SlipEternal
You are posting in a precalc forum, so I am not sure exactly what information you have to work with. Do you know that the function $e^x$ is a strictly increasing function? In other words, if $x then $e^x?
Yes, I know that. I'm only posting here because it's not a calculus question. But the root question wold go in the Calculus section

4. ## Re: Which direction for exponential inequality?

Yes, I know that. I'm only posting here because it's not a calculus question. But the root question wold go in the Calculus section
Excellent. I only meant that I would expect a calculus student to know that $e^x$ is a strictly increasing function. So, here is the explanation:

By the rules of logarithms, $x = e^{\ln x}$. Hence, you are trying to compare $e^{\ln x}$ and $e^y$. Since $e^x$ is a strictly increasing function and $x_1 = \ln x < y = x_2$, then $x = e^{\ln x} = e^{x_1} < e^{x_2} = e^y$.

5. ## Re: Which direction for exponential inequality?

x is less than e^(y) as
we can write ln(x) as(1/ log e base to x)
now cross multiply and u get (i have done for positive no assuming x>=e)(u can do the same for negative nos)
y*(log e base to x)>1
therefore
log (e^(y) base to x) >1 by using exponent property of (m*log b =log b^(m))
so as we need the log greater than 1
e^(y) should be greater than the base which is x
hence e^(y)>x
or x<e^(y)

6. ## Re: Which direction for exponential inequality?

Originally Posted by prasum
x is less than e^(y) as
we can write ln(x) as(1/ log e base to x)
now cross multiply and u get (i have done for positive no assuming x>=e)(u can do the same for negative nos)
y*(log e base to x)>1
therefore
log (e^(y) base to x) >1 by using exponent property of (m*log b =log b^(m))
so as we need the log greater than 1
e^(y) should be greater than the base which is x
hence e^(y)>x
or x<e^(y)
Thanks. But you broached another longstanding question I've never bothered to look up (which, now that I realize it, is related). Namely, if $\frac ab < \frac cd$, then which way does the inequality go when you cross-multiply and get $ad \quad bc$. (I'm going to guess that the answer depends on the signs of a, b, c, and d. But what if they're all the same sign?)

Originally Posted by SlipEternal
By the rules of logarithms, $x = e^{\ln x}$. Hence, you are trying to compare $e^{\ln x}$ and $e^y$. Since $e^x$ is a strictly increasing function and $x_1 = \ln x < y = x_2$, then $x = e^{\ln x} = e^{x_1} < e^{x_2} = e^y$.
TY. Understood everything.

7. ## Re: Which direction for exponential inequality?

Thanks. But you broached another longstanding question I've never bothered to look up (which, now that I realize it, is related). Namely, if $\frac ab < \frac cd$, then which way does the inequality go when you cross-multiply and get $ad \quad bc$. (I'm going to guess that the answer depends on the signs of a, b, c, and d. But what if they're all the same sign?)

TY. Understood everything.
Cross multiplying is a short cut that sometimes hides what is going on.

To keep things simple let's assume that either (1) a and c are both non-negative and b and d are both positive, or (2) a and c are both non-positive and b and d are both negative.

Consequently, b * d is positive.

$\dfrac{a}{b} < \dfrac{c}{d} \implies b * d * \dfrac{a}{b} < b * d * \dfrac{c}{d} \implies ad * \dfrac{b}{b} < bc * \dfrac{d}{d} \implies ad * 1 < bc * 1 \implies ad < bc.$

I'll let you work out what happens when the signs are not so neatly arranged.

8. ## Re: Which direction for exponential inequality?

No matter what the sign of a,b,c and d, we have:

$\dfrac{a}{b} < \dfrac{c}{d} \implies \dfrac{a}{b}\cdot\dfrac{d}{d} < \dfrac{c}{d}\cdot\dfrac{b}{b} \implies \dfrac{ad}{bd} < \dfrac{bc}{bd}$

If $bd > 0$ this means that $ad < bc$.

If $bd < 0$, we can multiply through by $|bd| = -bd$ (which is positive) to get:

$-ac < -bd$ so that $ac > bd$.

For example, 1/(-2) < 1/3, and we find that (1)(3) > (-2)(1). If we assumed $ac < bd$ no matter what the sign, this would be a counter-example.

For this reason, it is customary to put the negative sign(s) on the numerator(s) (if there is one/ there are any), so that we can always assume $bd > 0$.