Question #db5cd

Question

1189 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-10T16:37:16

#1 - lnroot(3)(2) = x#

Use #a^-n = 1/a^n#:

#1/e = 2e^(3x - 4)#

Cross multiply:

#1 = 2e^(3x - 4)e^1#

Use #a^m * a^n = a^(m + n)#

#1 = 2e^(3x - 4 + 1)#

#1 = 2e^(3x - 3)#

#1/2 = e^(3x - 3)#

Take the natural logarithm of both sides.

#ln(1/2) = ln(e^(3x- 3))#

Use #lna^n = nlna#.

#ln(1/2) = (3x- 3)lne#

#ln# and #e# are inverses--their product is #1#

#ln(1/2) = 3x- 3#

#1/3(ln(1/2) + 3) = x#

Use #ln(a/b) = lna - lnb#.

#1/3(ln1 - ln2 + 3) = x#

We know that #ln1= 0#.

#1/3(3 - ln2) = x#

#1 - 1/3ln2 = x#

Use #alnn = lnn^a#.

#1 - ln2^(1/3) = x#

#1 - lnroot(3)(2) = x#

Hopefully this helps!