logarithms

**BlueStar** · July 7th 2007, 09:51 AM

I'm having a real problem understanding logarithmic equations. Especially a certain area where you have to approximate values with given logarithms. I don't even really understand what I'm talking about.

For example:

If

find the approximate value of log3 8.

I've only been shown how to solve a problem like this that has an 'approximate value' that's a convenient multiple of the product of the two logarithms. Or one of the two logarithms can square itself and then conveniently go into it. I really don't get how to solve this one.

Also, I'm trying to figure out how to do this one:

Solve 10x = 152x.

log 10x = log 152x
x log 10 = 2x log 15

??

**Jhevon** · July 7th 2007, 09:56 AM

Originally Posted by BlueStar

I'm having a real problem understanding logarithmic equations. Especially a certain area where you have to approximate values with given logarithms. I don't even really understand what I'm talking about.

For example:

If

find the approximate value of log3 8.

Recall that $\log_a x^n = n \log_a x$

So we have $\log_3 8 = \log_3 2^3 = 3 \log_3 2 \approx 3(0.6309) = 1.8927$

Also, I'm trying to figure out how to do this one:

Solve 10x = 152x.

log 10x = log 152x
x log 10 = 2x log 15

??

I suppose you mean $10^x = 152^x$ ? use "^" to mean power
we would not use logs for this question. double check to see if you have a typo

**Jhevon** · July 7th 2007, 10:21 AM

For the record. Logarithms are exponents.

Definition: The Logarithm of a number to a given base is the power to which the base must be raised to give the number.

That is, if $\log_a b = c$ then $a^c = b$ .

example, what is $\log_2 8$ ?

$\log_2 8 = 3$

why? because we have to raise the base 2 to the power 3 to get the number 8

So logarithms are the exponents we use when we want to work in a particular base. Why all the fuss? Why not just use regular exponential equations to solve problems like these.

Straight-forward-matter-of-factly answer: try solving $3^x = 10$ using the techniques for regular exponential equations.

Better-more-wishful-thinking answer: Logarithms actually make life easier for the Mathematician. They can turn powers into multiples, multiples into additions, divisions into subtractions ...

The general log rules are self-explanatory of the properties of logarithms.

Rules:
Besides the definition, here are the three rules about logarithms that you MUST know.

$\log_a x^n = n \log_a x$

$\log_a xy = \log_a x + \log_a y$

$\log_a \left( \frac {x}{y} \right) = \log_a x - \log_a y$

Note: $\ln x \equiv \log_e x$ , where $e$ is Euler's constant.

Here are some rules that it's not that important for you to memorize, but they make life a whole lot easier if you do.

$\log_a 1 = 0$ , where $a > 1$ (I think we can have negative basis, but I've never worked with them, so I won't discuss them).

$\log_a 0$ does not exist

$\log_a a = 1$

$a^{\log_a x} = x$

$\log_a b = \frac {\log_c b}{ \log_c a}$ ----This is called "The change of base formula (for logarithms)"

$\log_a b = \frac {1}{ \log_b a}$

... and there are a few more that i'm forgetting at the moment. Like I said, it's not imperative that you memorize these (since they can be derived from the first four laws), but it's good if you do. Particularly the change of base formula, that shows up pretty often.

EDIT: I remembered two more rules. I added them

**BlueStar** · July 7th 2007, 02:29 PM

Okay, I'm starting to figure this out. So the log5 in the problem is superfluous?

**Jhevon** · July 7th 2007, 02:32 PM

Originally Posted by BlueStar

Okay, I'm starting to figure this out. So the log5 in the problem is superfluous?

yes, we didn't need $\log_3 5$ for that problem.

maybe there was another part to the question that we needed that for

**BlueStar** · July 7th 2007, 02:42 PM

Nope, that actually answered the question. Thank you.

Originally Posted by Jhevon

I suppose you mean $10^x = 152^x$ ? use "^" to mean power
we would not use logs for this question. double check to see if you have a typo

It would read like 10^x = 15^2x. Does that change it a little?

**Jhevon** · July 7th 2007, 02:48 PM

Originally Posted by BlueStar

It would read like 10^x = 15^2x. Does that change it a little?

That changes it a lot!

$10^x = 15^{2x}$

$\Rightarrow \log 10^x = \log 15^{2x}$ .......note, i mean log to the base 10

$\Rightarrow x \log 10 = 2x \log 15$

$\Rightarrow x = 2x \log 15$

$\Rightarrow 2x \log 15 - x = 0$

$\Rightarrow x \left( 2 \log 15 - 1 \right) = 0$

Can you take it from here?

**BlueStar** · July 7th 2007, 03:02 PM

How did you change that last x to a 1? The multiple x's were what was bothering me.

**Jhevon** · July 7th 2007, 03:03 PM

Originally Posted by BlueStar

How did you change that last x to a 1? The multiple x's were what was bothering me.

i factored a common x out. both terms have multiples of x in them

**BlueStar** · July 7th 2007, 03:08 PM

I don't get how that worked out...sorry. I've never seen a maneuver like that before. (as in changing an x to a 1)

**Jhevon** · July 7th 2007, 03:09 PM

Originally Posted by BlueStar

I don't get how that worked out...sorry. I've never seen a maneuver like that before. (as in changing an x to a 1)

really? i find that hard to believe, if you're doing logarithms.

Question: how would you solve for x here:

$x^2 - x = 0$

**BlueStar** · July 7th 2007, 03:13 PM

Originally Posted by Jhevon

really? i find that hard to believe, if you're doing logarithms.

Question: how would you solve for x here:

$x^2 - x = 0$

I don't really know......normally I'd think of isolating the first x by adding the other x to the other side....but that isn't right. I don't know how you would solve for x if there are only x's and no real numbers to compare to. I've actually been trying to breeze through a couple logarithm chapters and it's no wonder I'm in the dark about how to do these problems, and what I'm doing.

**Jhevon** · July 7th 2007, 03:26 PM

Originally Posted by BlueStar

I don't really know......normally I'd think of isolating the first x by adding the other x to the other side....but that isn't right. I don't know how you would solve for x if there are only x's and no real numbers to compare to. I've actually been trying to breeze through a couple logarithm chapters and it's no wonder I'm in the dark about how to do these problems, and what I'm doing.

well, doing logarithms is going to be hard if you don't know basic algebra, including factoring.

We can simply take the common term out, it's the reverse of expanding brackets.

$x^2 - x = 0$

$\Rightarrow x(x - 1) = 0$

Note that if you expand those brackets, you will get the original equation. From here we make the observation, that if two numbers being multiplied gives zero, it means that one number or the other is zero. So,

either $x = 0$ or $x - 1 = 0$

Thus, $\boxed { x = 0 }$ or $\boxed { x = 1 }$

How does factoring work exactly?

we search the terms to see what is common among them. In the case of $x^2 - x = 0$ it is clear that the common term is $x$ . now we simply put brackets around the expression and divide everything inside by the common term, while simultaneously multiplying the outside bracket by the common term.

$x \left( \frac {x^2}{x} - \frac {x}{x} \right) = 0$

since we are dividing and multiplying by the same thing, we are not changing anything, since this is essentially multiplying by 1. so now, we simplify to get

$x (x - 1) = 0$

and we continue as i described above.

i did the same thing with the log equation.

we had: $2x \log 15 - x = 0$

it wasn't hard to see that we have $x$ 's common to both terms on the left. we have $x$ multiplying $2 \log 15$ and $x$ multiplying -1

so now we pull out the common $x$ while simultaneously dividing by $x$ , we get:

$x \left( \frac {2x \log 15}{x} - \frac {x}{x} \right) = 0$

the $x$ 's in the bottom of the fractions cancel with the ones at the top and we have:

$x \left( 2 \log 15 - 1 \right) = 0$

now, since we have two numbers being multiplied gives zero, one number or the other is zero.

so, $x = 0$ or $2 \log 15 - 1 = 0$

since $2 \log 15 - 1 \neq 0$ we must have $\boxed {x = 0}$

**BlueStar** · July 7th 2007, 03:36 PM

I see it now. I understand factoring, just didn't think of it when I saw your equation. I'll need to memorize the steps.

**BlueStar** · July 7th 2007, 07:24 PM

Okay, now I just have a problem with this base e logarithm problem.

8 + 3e^3x = 26 (subtract 8)
3e^3x = 18 (divide the 3)
e^3x = 6 (divide the other 3?)
e^x = 2 (inverse property of equality)
x = ln2

x = .6931

The actual answer is .5973 and I can't seem to find out what I did wrong. Any help, please?

Thread: logarithms

LinkBack

Thread Tools

Display

logarithms

Similar Math Help Forum Discussions

logarithms help

Logarithms

logarithms

Logarithms

logarithms

Search Tags