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 is a strictly increasing function? In other words, if then ?
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 , then which way does the inequality go when you cross-multiply and get . (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.
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 $bd > 0$.