Thread: I'm stuck on logarithmic equations :(

I had a 98 in my Algebra class until this chapter came up. It's the last math course I ever have to take so I am hoping to finish strong.

Typically when a specific multiplier or divisor or whatever is present in every part of an equation, it can simply be removed, right? So if I have this equation:

log24x + log29 = log218

Am I allowed to simply remove the log portion from the equation and leave it was 4x+9=18? It just doesn't seem right. If not how do I solve it?

Is it the same with natural logs, since they always have the same base? Take

lnx- ln(x-4) = ln3

Thanks in advance.

Hi!

I would suggest that you take a good look at the logharitmic laws, and exponent laws as well.
You simply must now then well enough in order to solve these types of equations.

log2 (4x) = log 2 (4) + log2 (x)

These are one of the laws.

Another example: lg (10/3) = lg (10) - lg(3) = 1 - lg(3)

This is the second law.

I really really think you´ll easy crack these once you´ve mastered this laws.

Originally Posted by gregorywb

I had a 98 in my Algebra class until this chapter came up. It's the last math course I ever have to take so I am hoping to finish strong.

Typically when a specific multiplier or divisor or whatever is present in every part of an equation, it can simply be removed, right? So if I have this equation:

log24x + log29 = log218

Am I allowed to simply remove the log portion from the equation and leave it was 4x+9=18? It just doesn't seem right. If not how do I solve it?

Since the logarithms have the same base, we can combine them:

Thus, .

Now we can let the logs drop out of the equation to get

Originally Posted by gregorywb

Is it the same with natural logs, since they always have the same base? Take

lnx- ln(x-4) = ln3

Thanks in advance.

Yes. Since you are subtracting natural logarithms:

Thus, .

Now the ln(s) can drop out and we are left with

Can you take it from here?

I hope that this makes more sense now!

--Chris