7^0.3x = 813, take ln of both sides

(0.3x)(ln7) = ln(813)

0.3x = ln(813)/ln7

x = ln(813) / (0.3 * ln7)

Okay, I get the ln813, but why did you break up the left part like that?

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7^0.3x = 813, take ln of both sides

(0.3x)(ln7) = ln(813)

0.3x = ln(813)/ln7

x = ln(813) / (0.3 * ln7)

Okay, I get the ln813, but why did you break up the left part like that?

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To equate each side of the equation. If you make a change to one side of the equation, you must make an appropriate alteration to the other, otherwise the entire thing becomes inherently changed from the initial problem you were trying to solve.Okay, I get the ln813, but why did you break up the left part like that?

Don't I know it...We all make simple mistakes.

I know, I was just trying to show a few more steps after reading her post about not understanding.And both of these are exactly the same as my own, expressed in a slightly different form.

That's exactly right.Okay, on the third problem after the -1/6 problem, I think I'm beginning to understand. ln and the e's cross out, leaving 1-5x=ln793. The next thing you do is isolate the x, by gettind rid of the 1. 5x=ln793-1. Then, divide by 5 to get the x alone some more. There's the answer. Is that right, is that how it's done?

The reason e and ln cancel each other out, is because they are inverse functions of each other. If you have a^x = a, we can turn this into log_a(a) [That's log base a, of a...]. If the base of the logarithm matches the value you're taking the logarithm of, they cancel and give you one (ie log_2(2) = 1, log_36x(36x) = 1). Well, ln is just a special case of log, in that ln's base is always e. So ln(e) = log_e(e) = 1.

To equate each side of the equation. If you make a change to one side of the equation, you must make an appropriate alteration to the other, otherwise the entire thing becomes inherently changed from the initial problem you were trying to solve.

So why didn't you put the ln with 0.3x? You added ln to the 7. I think I get the rest. You isolate the x, so you divide both sides by ln 7. But on the last step why do you mutiply the power by ln 7?

My book and teacher keep driving me in circles on this.

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It describes the ways in which electrons are organised around the atom in various orbits. The lowest level can store 2 electrons, the next one up 6, then 10, 14, 18, and more, but you probably won't go into those extras too much. One orbit doesn't need to be 'filled up' for the next orbit up to contain electrons. In a normal atom, the number of electrons equals the number of protons, giving the atom equal positive and negative charges, making it neutral. Removing or adding atoms to an atom changes its electric charge and ionises it. Some of the basics.

My book and teacher keep driving me in circles on this.

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OK im still confused on how this works around the elements since at times for me it contradicts with other charts for me, also can someone tell me how this affects waves of atoms?(its a bonus question my teacher promised to put :3)It describes the ways in which electrons are organised around the atom in various orbits. The lowest level can store 2 electrons, the next one up 6, then 10, 14, 18, and more, but you probably won't go into those extras too much. One orbit doesn't need to be 'filled up' for the next orbit up to contain electrons. In a normal atom, the number of electrons equals the number of protons, giving the atom equal positive and negative charges, making it neutral. Removing or adding atoms to an atom changes its electric charge and ionises it. Some of the basics.

Instructions:

Use Log properties to expand log expressions as much as possible, if possible, evaluate without the use of a calculator.

Log18^x-3^11

ln(e^2x over X square root)

Thank you.

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It's the number of protons that define an element mostly. A chunk of matter with five protons and theoretically any number of neutrons and electrons is still boron. Changing the number of electrons alters its charge and electromagnetic properties, while a change in neutrons produces an isotope which is more of a structural change.OK im still confused on how this works around the elements since at times for me it contradicts with other charts for me,

Matter can demonstrate wavelike properties, but at low speeds these are virtually unnoticeable. Accelerate something small to high speeds and it'll start assuming wave properties. I'm not sure how electrons alone might affect the whole, but in this case we're talking about the de Broglie wavelength of a subject and matter waves. I don't think this directly answers your bonus question, but it might help point you in the right direction.also can someone tell me how this affects waves of atoms?(its a bonus question my teacher promised to put :3)

This would require the use of the logarithm laws in particular, among general maths skills.Instructions:

Use Log properties to expand log expressions as much as possible, if possible, evaluate without the use of a calculator.

log

log

log

log

Wish I had time to help with more right now.

log[/B]a(m^n) = n*loga(m)a(

loga) = 1a

log(1) = 0

Wish I had time to help with more right now.

Okay, so when you mutiply powers, your break them apart by adding them.

When you divide powers, you break them apart by subtraction.

So, on the third one, is it when you have the argument raised to a power, you move that power to the front?

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Yeah, raising the argument to the power ofSo, on the third one, is it when you have the argument raised to a power, you move that power to the front?

16 ^ (x-9) = 3 ^ (-9x)

or

any help on this would be greatly appreciated. I know you need to use logarithms to solve this equation, but I don't exactly know how to go about doing so. I've tried a few different ways, and I get a different answer each time.

oh, and if it's any help, the answer is asked for in base-10 logarithms.

ln both sides. That's all I know, I don't know where you'd go from there.

I can't "ln" both sides, because it has to be log base-10, not natural log :/

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I'm sure this comes out to be something algebraically simple, but I'm not sure how to piece it together. I'd imagine that it's a system of equations with the first being x + (5/2)y + z = 500,000, but I'm not sure how to set up the equation for the annual interest rate. I'm not sure if that means the average interest rate or whatever else, since the rate keeps changing. Any ideas?

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Okay, let's say:

y = amount borrowed at 10%

z = amount borrowed at 12%

Therefore:

I'm assuming we're talking about the first year of interest (or simple interest) here. The total interest will be 9% of amount x, plus 10% of amount y, plus 12% of amount z:

The amount borrowed at 10% is y; the amount borrowed at 9% is x. Therefore:

You now have three equations and three unknowns, all you need to do is solve them simultaneously. I'd start by substituting (iii) into (ii) and (i).

16 ^ (x-9) = 3 ^ (-9x)

or

any help on this would be greatly appreciated. I know you need to use logarithms to solve this equation, but I don't exactly know how to go about doing so. I've tried a few different ways, and I get a different answer each time.

oh, and if it's any help, the answer is asked for in base-10 logarithms.

I see it hasn't been solved yet, i can help. You have to take the log with base ten of both sides, although this doesn't seem to help, it does as you must use a certain property logarithms have. The property is as follows,

if you take ln(a^2), it equals to 2*ln(a).

For an explanation of this, see this link. Using this property you will come to the following equation

(x-9)*log(16)=-9x*log(3)

after which you can solve for x using basic algebra.

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I've been looking around the internet, and i have yet to found any good example

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Dr Math said:Complex numbers enter into studies of physical phenonomena in ways

that most people can't imagine. There is, for example, a differential

equation, with coefficients like the a, b, and c in the quadratic

formula, that models how electrical circuits or forced spring/damper

systems behave. The movement of the shock absorber of a car as it goes

over a bump is an example of the latter. The behavior of the

differential equations depends upon whether the roots of a certain

quadratic are complex or real. If they are complex, then certain

behaviors can be expected. These are often just the solutions that one

wants.

In modeling the flow of a fluid around various obstacles, like around

a pipe, complex analysis is very valuable for transforming the problem

into a much simpler problem.

When everything from large structures of riveted beams to economic

systems are analyzed for resilience, some very large matrices are used

in the modeling. The matrices have what are called eigenvalues and

eigenvectors. The character of the eigenvalues, whether real or

complex, is important in the analysis of such systems.

In everyday use, industrial and university computers spend some

fraction of their time solving polynomial equations. The roots of such

equations are of interest, whether they are real or complex.

And complex numbers are useful in studying number theory, which is the

study of the positive integers. The techniques in complex analysis

are just one more tool that researchers have.

Math Forum - Ask Dr. Math

Neatly organized into five paragraphs, too.

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