i've been trying to figure this one out all night...?

Question

solve each inequality:

a) 1 < ln x < 6

b) e^(2-3x) > 5

Answer 1

Easiest way to solve these is to either take the inverse natural log (ie e^(that value)) or take the natural log so that the equation has values and x is by itself

e^(ln y) = y
ln(e^y) = y

a) 1< ln x < 6 (to get x as a singular value put each side of the eqn as the power of e)

e^1 < e^(ln x) < e^6
2.718 < x < 403.429

b) e^(2-3x) > 5
ln(e^(2-3x)) > ln(5)

2-3x > 1.609
0.391 > 3x
0.391/3 > x
0.130 > x

Hope this helps

Answer 2

Why, they aren't all that hard?

a: You can cast each term as an exponent of "e" and maintain the inequality. Then you have ....
e< x < e^6

b: You can take the natural log of each term and maintain the inequality. Then you have....
2-3x > ln 5, which can be spiffed up to get a result in x: 3x-2 < ln 5 [sneaky!] and
x < (2 - ln 5)/3

Answer 3

the inverse function for the natural log is the exponential function, e, so to solve the first one, you take the e to everything:

e^1 < e^(ln(x)) < e^6

so, since e and ln are inverses they "cancel out" to leave x, e^1 is of course e (a constant), so the answer is of course

e < x < e^6

Similarly, with the second question, you take the natural log of it all

ln(e^(2-3x)) > ln(5), since e and ln are inverses, they leave 2-3x

2 - 3x > ln(3x) and now we solve for x

x < -(ln(3x)-2)/3, don't forget to switch the sign of the inequality when you divide by a negative number!

Answer 4

is this a true or false questions ?