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How do you solve #ln (x – 2) + ln (x + 2) = ln 5#?

Answer 1

Chloe Alexander

Use properties of #ln# to find #ln(x^2-4) = ln(5)# and hence find solution: #x = 3#

#ln(x^2-4) = ln((x-2)(x+2)) = ln(x-2)+ln(x+2)#

#= ln(5)#

So #x^2-4 = 5#

Add #4# to both sides to get: #x^2 = 9#

So #x = +-3#

If #x = -3#, then #x - 2 < 0# and #x + 2 < 0#, so neither #ln(x-2)# nor #ln(x+2)# have Real values.

If #x = 3# then:

#ln(x-2) + ln(x+2) = ln(1) + ln(5) = 0 + ln(5) = ln(5)#

So the only solution is #x = 3#

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Answer 2

Ariana Alvarez

Well you can write it as follows

#ln(x-2)+ln(x+2)=ln5=>ln(x^2-4)=ln5=>x^2-4=5=>x^2=9=>x=-3,x=+3#

But x-2>0 so the solution is x=3

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Answer 3

Greyson Brown

To solve the equation ln(x - 2) + ln(x + 2) = ln(5), you can combine the logarithms using the property of logarithms that states ln(a) + ln(b) = ln(a * b). So, ln((x - 2)(x + 2)) = ln(5). This simplifies to ln(x^2 - 4) = ln(5). Then, by equating the arguments, you get x^2 - 4 = 5. Solve for x to find x = ±√9. Therefore, x = ±3. However, since the natural logarithm ln is only defined for positive real numbers, the solution is x = 3.

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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.