Solving natural logarithm problems

[outlawchesus]

3Ln 5x= 10


Ln(x+5) = Ln(x-1) - Ln(x+1)

do you add e to each side and then go from there, if thats right what do you with the 3?

second one:

x+5/x-1=x+1 would get rid of each of the natural logs and then just cross multiply?


i'm pretty insecure with my math...

 

[DrMike]

For any log question, try to remember what logs actually mean.

log[sub:1r404t8d]b[/sub:1r404t8d]a=c just means a=b[sup:1r404t8d]c[/sup:1r404t8d]

So,

log[sub:1r404t8d]2[/sub:1r404t8d]32=5 just means 32=2[sup:1r404t8d]5[/sup:1r404t8d]

Also,

ln x just means log[sub:1r404t8d]e[/sub:1r404t8d]x, where e is the crazy or beautiful number 2.7182818284... There are very good reasons why e is called the "natural" base of for logarithms, but you need a bit of calculus to see them.

When you say "do you just add e to each side"...

if you mean

"change ln(x+5) = ln(x-1) - ln(x+1) to log[sub:1r404t8d]e[/sub:1r404t8d](x+5) = log[sub:1r404t8d]e[/sub:1r404t8d](x-1) - log[sub:1r404t8d]e[/sub:1r404t8d](x+1)",

then, "yes", go for it!

Then, you need to remember your log laws - all derived from the fact that "log[sub:1r404t8d]b[/sub:1r404t8d]a=c means a=b[sup:1r404t8d]c[/sup:1r404t8d]"

eg "what do you do with the 3?"

You could remember that

d log[sub:1r404t8d]b[/sub:1r404t8d]a = log[sub:1r404t8d]b[/sub:1r404t8d](a[sup:1r404t8d]d[/sup:1r404t8d]).

This is true because
if log[sub:1r404t8d]b[/sub:1r404t8d](a[sup:1r404t8d]d[/sup:1r404t8d])=x,
it means a[sup:1r404t8d]d[/sup:1r404t8d] = b[sup:1r404t8d]x[/sup:1r404t8d].
and if log[sub:1r404t8d]b[/sub:1r404t8d](a)=y,
it means a = b[sup:1r404t8d]y[/sup:1r404t8d].
But if a = b[sup:1r404t8d]y[/sup:1r404t8d] and a[sup:1r404t8d]d[/sup:1r404t8d] = b[sup:1r404t8d]x[/sup:1r404t8d],
then (b[sup:1r404t8d]y[/sup:1r404t8d])[sup:1r404t8d]d[/sup:1r404t8d]=b[sup:1r404t8d]x[/sup:1r404t8d], so
b[sup:1r404t8d]dy[/sup:1r404t8d] = b[sup:1r404t8d]x[/sup:1r404t8d].
But this means that dy=x.
Therefore, d log[sub:1r404t8d]b[/sub:1r404t8d]a = log[sub:1r404t8d]b[/sub:1r404t8d](a[sup:1r404t8d]d[/sup:1r404t8d]).

Also, your Ln(x+5) = Ln(x-1) - Ln(x+1) doesn't become
x+5/x-1=x+1
or even (what I think you meant)
(x+5)/(x-1)=x+1

You have the right idea, but I think you made a mistake somewhere. Do it again carefully....

log(b) - log(c) = log(b/c),
so if log(a) = log(b)-log(c), then log(a) = log(b/c), so a = b/c. Then, as you said, just cross-multiply and solve the quadratic equations....

 

[mmm4444bot]

I do not understand your work on the first exercise. It looks like you might have started (on the left-hand side) by trying to use the property of logarithms that deals with a sum of logarithms; if this is so, then it's not quite correct.

[In general, the expression ln(5x) does not equal the expression ln(x + 5). These two expressions are only equal when x is 5/4. But 5/4 is not the solution to your exercise.]

Here is the correct application of that property:

ln(5x) = ln(x) + ln(5)

This is true for ALL positive Real values of x.

The right-hand side of your work is a total mystery, to me. But, that's okay.

I would start this exercise using a different strategy. A logarithmic equation can be rewritten as an exponential equation.

Here's a similar example.



7 ln(9x) = 16

Isolate the natural log factor by dividing both sides by 7.

ln(9x) = 16/7

Remember that logarithms are exponents. Therefore, the above equation tells us that 16/7 is an exponent. Use this fact, along with the definition of the natural logarithm, to rewrite the above equation as an exponential equation.

e^(16/7) = 9x

Divide both sides by 9 to isolate the x factor.

e^(16/7) / 9 = x

Use a calculator to evaluate the left-hand side.

1.092523009 = x



We can check this result for x by substituting into the original logarithmic equation.

7 ln(9x) = 16

7 ln(9 * 1.092523009) = 16

Using a calculator, we get the following confirmation.

16 = 16

Your exercise may be solved using the same steps.

If I wrote anything that you do not understand, then please post specific questions, and I will clarify with more examples.

If you would like more help with this exercise, then please continue to show whatever work you can accomplish, and try to say something about why you're stuck, so that I might determine where to continue helping.