How do you solve #ln(x - 3) + ln(x + 4) = 1#?

Answer 1

#x~=3.37#

First thing to note is that a #ln(x-3)# is defined only when #x-3>0=>x>3#
And also #ln(x+4)# is defined when #x+4>0 => x> -4#
And so for both functions to be defined we need to intersect their domains, leading to : #x>3#

Now, that we have the domain we can solve the equation in this domain, thus:

#ln(x-3)+ln(x+4)=1#
We apply the logarithm laws #=> ln[(x-3)(x+4)]=1#
#=>(x-3)(x+4)=e#
#e# is Euler's constant
#=>x^2+x-12=e#
#=>x^2+x-12-e=0#

We apply the quadratic formula,

#=>x=(-1+-sqrt(1-4(-12-e)))/2=(-1+-sqrt(49+4e))/2#
The solution #(-1-sqrt(49+4e))/2# has to be rejected because it is negative and so falls out our domain #x>3#
The only solution is #x=(-1+sqrt(49+4e))/2~=color(blue)3.37#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To solve the equation ln(x - 3) + ln(x + 4) = 1, you can use the properties of logarithms to combine the two logarithmic terms into a single logarithm. Then, you can exponentiate both sides to eliminate the logarithm. After that, you can solve for x. Here are the steps:

  1. Combine the logarithmic terms using the property: ln(a) + ln(b) = ln(a * b). ln(x - 3) + ln(x + 4) = ln((x - 3)(x + 4)).

  2. Rewrite the equation: ln((x - 3)(x + 4)) = 1.

  3. Exponentiate both sides using the property of logarithms: e^ln(u) = u. e^ln((x - 3)(x + 4)) = e^1.

  4. Simplify: (x - 3)(x + 4) = e.

  5. Expand and rearrange the equation: x^2 + x - 12 - e = 0.

  6. Solve the quadratic equation for x using the quadratic formula or factoring.

  7. Once you find the solutions for x, check if they are valid by ensuring they satisfy the domain of the original logarithmic equation, which requires x > 3 and x > -4.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7