How do you solve this equation? #-aln(a-x)=ln(t-b)#
#-aln(a-x)=ln(t-b)#
Solve for x.
Solve for x.
1 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
Next, we need to split into cases
Case
In this case we have
#0 = ln(t-b)#
So
#t - b = 1#
Adding
#t = b + 1#
The value of
#a - x > 0#
Hence
#x < 0#
Case
With
#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
#e^ln(r) = r#
Also if
#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
#1/(t-b) > 0#
and the principal
#root(a)(1/(t-b))#
is a well-defined positive Real number.