Logarithm help!

Please help!

1) 2 - logsub5 (25x)

2) ln (e^2x) + ln (ey) - 3

3)ln (x+8) - lnx = 1

any help would be great!!!

Please specify your question(s) more exact. What can we do with the first two equations? There isn't even any "=" sign so we can't solve it for x or anything...

As for the third, do you know that:

e^ln(x) = x
and
e^{ln(a) + ln(b)} = e^ln(a) * e^ln(b) = a * b
and
e^{ln(a) - ln(b)} = e^ln(a) / e^ln(b) = a / b

The same goes for log functions, just keep in mind that ln(x) = log_e(x)
So if you have a log₅ function instead of the ln function, you only have to replace the "e", which is the base of the ln function, by "5", which is the base for this particular log function (indicated by the subscript 5).

That will help you solve it. If not just ask.

sorry! the first one just says to make it into a single log.

Ok, and the second, make it into a single ln?

For the first, I'll give you the first step:

2 = log₅(5²)

This is true because you could say the log and the 5^.. cancel each other out.
In general:

log_a(a^x) = x

The same goes for ln:
ln(e^x) = x
Because ofcourse, ln(x) = log_e(x)


Combine that with the fact that

log₅(a) - log₅(b) = log₅(a / b)

and you can probably manage it!



Did you solve the third equation?

I find it a little helpful to think of logarithms as the inverse of the exponent operation. Addition has it's inverse--subtraction; same with multiplication and division. Just like in general you need to know what "base" you are exponentiating [4^2--you're exponentiating the base 4 by 2], you need to know what "base" you're using to invert that operation. So, you always need a "base" for the logarithm to work off of. Later on it turns out that e^x and its inverse, lnx, are very useful, so the base "e" is introduced very early even if it seems magical.

In that light the logarithm rules make a bit more sense intuitively.

log_b(x*y) = log_b(x) + log_b(y)

parallels

b^x+y = b^x*b^y,

but notice that it's reversed--the addition occurs inside the exponent instead of between exponents, while the addition occurs between the logarithms instead of inside the logarithm.

The mystical property that e^lnx = x is also obvious from viewing these as inverse functions. You can make similar parallels to the other properties of the logarithm. You can also understand why the argument to the logarithm has to be greater than zero--because the graph of exponential functions, when inverted by flipping across the x=y line, start at -infinity around 0. That is, since 10^-infinity goes to 0, log₁₀0 goes to -infinity. Similarly, since e^x must be greater than 0 always, the natural log must accept an argument greater than zero always (though since x can be negative, ln can return a negative number in some cases).

Anywho, after you conceptually understand the logarithm (if you're interested enough to do so), the best idea is to simply memorize and use the rules of logarithms if you ever need to become fluent in them. And if you plan on taking higher level math, science, engineering, etc. courses, you'll have to be familiar with them.

Indeed, the rules for logs are sometimes very hard to understand and may seem arbitrary at first, I had the same feeling when I saw the rules for the first time.

However, after you use them a few times they tend to start to feel more natural; just like everyone knows that substraction is the inverse of addition, after a while you will also simply know that a log is the inverse of an exponent.