How do you solve #ln(x-2)-ln(x+2)=2#?

2 Answers
Jul 28, 2016

Answer:

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

Explanation:

This problem requires some clever manipulation and knowledge of logarithm/exponential identities but is easily solved with the right toolset.

We start by taking the exponential function of both sides. That is, replacing each side with #e# to the power of that side:

#e^(ln(x-2) - ln(x+2)) = e^2#

Now we use the fact that subtraction in an exponential is the same as division outside it:

#e^(ln(x-2) - ln(x+2)) = e^(ln(x-2))/e^(ln(x+2)) = e^2#

Now we use the fact that the exponent of the natual logarithm of any number is just that number:

#e^(ln(x-2))/e^(ln(x+2)) = (x-2)/(x+2) = e^2#

From here, the solution proceeds by standard algebra:

#(x-2)/(x+2) = e^2#

#x-2 = e^2(x+2)#

#x - 2 = e^2 x + 2e^2#

#x - e^2 x = 2e^2 + 2#

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

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

Jul 28, 2016

Answer:

No solution, if we are only considering Real values.

Upon examining Complex solutions, we find that #x=(2(e^2+1))/(1-e^2)#.

Explanation:

From the outset, we can apply the logarithm rule:

#color(red)(barul|color(white)(a/a)color(black)(log(A)-log(B)=log(A/B))color(white)(a/a)|)#

Yielding the simpler equation:

#ln((x-2)/(x+2))=2#

From here, use the relation:

From the outset, we can apply the logarithm rule:

#color(blue)(barul|color(white)(a/a)color(black)(ln(A)=B" "hArr" "A=e^B)color(white)(a/a)|)#

This can also be viewed as exponentiating both sides with a base of #e#.

Thus:

#(x-2)/(x+2)=e^2#

Using the algebra already outlined in the other answer:

#x-2=e^2(x+2)#

Distributing:

#x-2=e^2x+2e^2#

Grouping the #x# terms:

#x-e^2x=2e^2+2#

Factoring:

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

Dividing:

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

IMPORTANT NOTE!

The value of #(2(e^2+1))/(1-e^2)# is approximately #-2.6261#. Thus, in #ln(x-2)# and #ln(x+2)#, the argument of the natural logarithm is negative. For the typical Real-valued logarithm function, this is prohibited.

Thus the value of #x=(2(e^2+1))/(1-e^2)# is only allowed if we are assuming to be working with a Complex-valued logarithm.