Solve Natural Logarithms ln(x+1)+ln(x+3) < ln(x+7)

In summary, to find x such that ln(x+1) + ln(x+3) < ln(x+7), we first solve ln((x+1)(x+3)) = ln(x+7), which simplifies to (x+1)(x+3) = x+7. This is a quadratic equation that can be solved using various methods such as factoring, the quadratic formula, or completing the square. The solutions to this equation are x = -14 and x = 11. However, we must also consider the restrictions of the logarithm function, which states that x must be greater than 0. Therefore, the final solution is x = 11.
  • #1
Compaq
23
0
ln(x+1) + ln(x+3) < ln(x+7)

Find x!


I should of course know this, second year at high school (in Norway). I've just forgotten the way attack the whole thing^^
 
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  • #2
ln(x+1)+ ln(x+3)= ln((x+1)(x+3)

To find x such that ln((x+1)(x+3))< ln(x+7), it is probably simplest to first solve
ln((x+1)(x+3))= ln(x+7) which, because ln is a "one-to-one" function, is equivalent to (x+1)(x+ 3)= x+7, a quadratic equation. The solutions to that equation separate the number line into intervals where ln(x+ 1)+ ln(x+3)< ln(x+7 for every point in the interval or is not true for every point.
 
  • #3
Ahh, I see. By factorising with one of the logarithm rules, I get the ln to one number equals the ln to another number, thus the numbers must be the same.

Thanks, this + other tasks are to be handed in on Monday:)

PS, not sure if this was put in the correct forum.

So, anyway. Those of you who answer our questions. Who, or what, are you? Students? Educated people just coming in here regularly to answer?

Regards, Compaq
 
  • #4
I got:

x = -14 V x = 11

Is that correct?
 
  • #5
Compaq said:
Ahh, I see. By factorising with one of the logarithm rules, I get the ln to one number equals the ln to another number, thus the numbers must be the same.

Thanks, this + other tasks are to be handed in on Monday:)

PS, not sure if this was put in the correct forum.

So, anyway. Those of you who answer our questions. Who, or what, are you? Students? Educated people just coming in here regularly to answer?

Regards, Compaq
Different people, different reasons for being here :smile: I often need help since some of my teachers don't know the answers to my problems (yes, even they are human), but I feel I need to give back to the community, so I help out where I can. Of course, ultimately we are all here to learn a thing or two.


Compaq said:
I got:

x = -14 V x = 11

Is that correct?
Um not quite.

where you left off: [tex](x+1)(x+3)<x+7[/tex]

expanding and collecting like terms: [tex]x^2+3x-4<0[/tex]

factorising: [tex](x+4)(x-1)<0[/tex]

Now from here, I find it easiest to graph the quadratic [tex]y=(x+4)(x-1)[/tex] and since y<0, the domain is for all values of x which y<0 (looking at the graph, wherever the quadratic is below the x-axis).

If you could follow that, you should find that [tex](x+4)(x-1)<0[/tex], then [tex]-4<x<1[/tex]

ok so it looks like it is answered, but because we manipulated the problem in such a way as to rid ourselves of the logarithm, there might be more restrictions on the value of x. Look back at the original question, for [tex]ln(x)[/tex] what can't the value of 'x' be?
 
  • #6
Yeah, I screwd up with the general forumula for quadrics (or whatever you call itXD), so I tried with that other way (which I don't know the English term of).. (I add (b/2)2 and then add -(b/2)2 ) I got the same factors as you, at least.

And we're dealing with the logarithm function, so x > 0, negative numbers, nor sero, have logarithms, logically. So:

(x+4)(x-1) < 0 when 0<x<1

Does that seem right, then:)

And thanks a lot:D

One more question, how do you add fractions and all that stuff, I feel I'm using a werid method here:P
 
  • #7
Ohh, and I always use factor tables...
 
  • #8
Wait! Of course it should be:

ln(x+1) + ln(x+3) < ln(x+7) when 0<x<1
 
  • #9
Compaq said:
Yeah, I screwd up with the general forumula for quadrics (or whatever you call itXD), so I tried with that other way (which I don't know the English term of).. (I add (b/2)2 and then add -(b/2)2 ) I got the same factors as you, at least.
Sure, do it however you like. Factoring, using the quadratic formula and completing the square all work.

And we're dealing with the logarithm function, so x > 0, negative numbers, nor sero, have logarithms, logically. So:

(x+4)(x-1) < 0 when 0<x<1

Does that seem right, then:)

And thanks a lot:D
Not exactly. For [tex]ln(a), a>0[/tex]
Which means in the question, [tex]ln(x+1) + ln(x+3) < ln(x+7)[/tex]

For each term inside the logarithm: [tex]x+1>0 , x+3>0 , x+7>0[/tex]

Simplifying: [tex]x>-1 , x>-3 , x>-7[/tex]

But all these inequalities of x must be satisfied, so take the largest: [tex]x>-1[/tex]

So what is the final result then? Remember: both the answer we got from finding the inequality [tex](x+4)(x-1)<0[/tex] and [tex]x>-1[/tex] must be satisfied.


One more question, how do you add fractions and all that stuff, I feel I'm using a werid method here:P
I'm not exactly sure what you're asking. Create another help thread :smile:

Ohh, and I always use factor tables...
Sorry I've never heard of factor tables, so I wouldn't know what they are.
 

What is a natural logarithm?

A natural logarithm, denoted as ln, is a mathematical function that is the inverse of the exponential function. It is used to find the exponent that a base number must be raised to in order to obtain a given value.

How do I solve natural logarithms?

To solve natural logarithms, you can use the properties of logarithms, such as the product, quotient, and power rules. In this case, you can combine the two logarithms using the product rule, and then use algebraic manipulation to isolate the variable on one side of the inequality.

What is the solution to ln(x+1)+ln(x+3) < ln(x+7)?

The solution to this inequality is x < 2. This can be found by combining the two logarithms using the product rule, and then solving for x using algebraic manipulation.

Why is it important to solve natural logarithms?

Solving natural logarithms is important in many areas of science, including physics, chemistry, and biology. It allows us to find the value of a variable that is hidden in an exponential function, which is often used to model many natural phenomena.

Are there any restrictions when solving natural logarithms?

Yes, there are a few restrictions to keep in mind when solving natural logarithms. First, the argument of the logarithm must be a positive number. Secondly, when using the power rule, the base of the logarithm must also be a positive number. Lastly, the solution must make sense in the context of the problem, and any extraneous solutions must be discarded.

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