How do you solve this equation? #-aln(a-x)=ln(t-b)#

#-aln(a-x)=ln(t-b)#
Solve for x.

1 Answer
Apr 17, 2016

Answer:

#x = a - root(a)(1/(t-b))" "# if #a != 0#

#x < 0" "# if #a = 0#

Explanation:

Your equation looks like this

#-a * ln(a-x) = ln(t-b)#

You're dealing with natural logs, so right from the start you know that you must have

#a -x > 0" "# and #" " t -b > 0" " " "color(orange)("(*)")#

since you can't take the natural log of zero or a negative number while working in #RR#.

Next, we need to split into cases #a = 0# and #a != 0#

Case #bb(a = 0)#

In this case we have

#0 = ln(t-b)#

So

#t - b = 1#

Adding #b# to both sides we find

#t = b + 1#

The value of #x# is unconstrained except that we require

#a - x > 0#

Hence

#x < 0#

Case #bb(a != 0)#

With #a != 0# we can first divide both sides of the equation by #-a# to get

#ln(a - x) = ln(t-b) * -1/a#

Then taking exponents of both sides

#e^(ln(a-x)) = e^(ln(t-b)*-1/a)#

Note that if #r > 0# then

#e^ln(r) = r#

Also if #A > 0# and #B, C in RR#, then

#A^(BC) = (A^B)^C#

So our equation simplifies to

#a-x = (t-b)^(-1/a)#

Hence

#x = a - (t-b)^(-1/a) = a - root(a)(1/(t-b))#

Note that since #t-b > 0# we also have

#1/(t-b) > 0#

and the principal #a#th root represented by

#root(a)(1/(t-b))#

is a well-defined positive Real number.