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

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

Solve for x.

Solve for x.

##### 1 Answer

#### 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.