How do you solve 5^(x+3) <= 2^(x+4)?

1 Answer
Aug 9, 2015

Replace 5 with e^(ln(5)) and 2 with e^(ln(2)); then solve using the new exponents (ignoring the identical base values):
color(white)("XXXX")x<= -2.24353

Explanation:

Let k = ln(5) =1.609438 and
let m = ln(2) = 0.693147

5 = e^kcolor(white)("XXXX")andcolor(white)("XXXX")2 = e^m

5^(x+3) <= 2^(x+4)
color(white)("XXXX")is equivalent to
(e^k)^(x+3) <= (e^m)^(x+4)

kx+3k <= mx+4m

(k-m)x <= (4m-3k)
color(white)("XXXX")since k>m we can divide both sides by k-m
color(white)("XXXX")without changing the orientation of the inequality

x <= (4m-3k)/(k-m)

Substituting the value of ln(5) for k
and the value of ln(2) for m
then pushing it all through my calculator...

I get x <= -2.24353