# Thread: I'm stuck on logarithmic equations :(

1. ## 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

2. ## hi

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.

3. 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: $\displaystyle \log_a(b)+\log_a(c)=\log_a(bc)$

Thus, $\displaystyle \log_2(4x)+\log_2(9)=\log_2(18)\implies \log_2(36x)=\log_2(18)$.

Now we can let the logs drop out of the equation to get $\displaystyle 36x=18\implies x=...$

Originally Posted by gregorywb
Is it the same with natural logs, since they always have the same base? Take

lnx- ln(x-4) = ln3

Yes. Since you are subtracting natural logarithms: $\displaystyle \ln(a)-\ln(b)=\ln\bigg(\frac{a}{b}\bigg)$
Thus, $\displaystyle \ln(x)-\ln(x-4)=\ln(3)\implies \ln\bigg(\frac{x}{x-4}\bigg)=\ln(3)$.
Now the ln(s) can drop out and we are left with $\displaystyle \frac{x}{x-4}=3$