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# Homework help... Again

Former Member
Posts: 1,876,328 The Mix Honorary Guru

Have a test tomorrow and going through worksheets which hold the same kind of questions as my test.

Have forgotten so much of the stuff from last year and I am seriously screwed with this.

Help to solve these questions would be dearly appericiated, you have no clue.

1.

If I get told to solve the equation: log(x+1) + log(x-2) = 1, then how the hell do I even start?

2.

How do I solve the equation: 3000=167952,882*x^-2,017?

3.

How do I solve the equation: 90=1117,74*x^-0,805?

So basically I am mostly screwed with log and polynomials/power (that's what the dictionary says it's called in English).

Please somebody help

Have forgotten so much of the stuff from last year and I am seriously screwed with this.

Help to solve these questions would be dearly appericiated, you have no clue.

1.

If I get told to solve the equation: log(x+1) + log(x-2) = 1, then how the hell do I even start?

2.

How do I solve the equation: 3000=167952,882*x^-2,017?

3.

How do I solve the equation: 90=1117,74*x^-0,805?

So basically I am mostly screwed with log and polynomials/power (that's what the dictionary says it's called in English).

Please somebody help

## Comments

Jaqu shall I entertain you till a genius arrives cos I got lost soon as you mentioned test

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0 ·exp{log(x+1) + log(x+2)} = exp {1}

Then use the laws of powers (exp{a+b} = exp{a}*exp{b})to separate out the left hand side:

exp{log(x+1)}*exp{log(x+2)} = exp {1}

Generally exp{log(y)} = y if y>0, so applying this rule:

(x+1)*(x+2) = exp {1} [note condition that x>2]

Then expand out the factors and solve the quadratic for x!

At least I think that lot is right

2 and 3. Remember this simple rule "whatever you do to one side of the equation, you must also do to the other. For 2:

3000=167952,882*x^-2,017

Add 2,017 to both sides:

3000 + 2,017 = 167952,882*x

Then divide both sides by 167952,882:

3002,017 / 167952,882 = x

And do a similar thing for 3.

Hope that helps

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0 ·I failed the assignment but no surprsies as I have no idea what I'm doing! Think I will need to google to find help (not cheat I mean helpful explanations as the uni text books are pants!)

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0 ·:eek2:

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0 ·So I'll use this one instead:crazyeyes

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0 ·Bopz :crazyeyes

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0 ·Though I don't think that the last calculation is correct. Don't have the strength to point it out now. Still have a lot of math to go through before tomorrow :crazyeyes

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0 ·Depends, I aint that hot on stats!

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0 ·Nuts, you're quite right, I missed the fact that they are raised to a power!

Just ignore the addition bit....divide both sides by the 167952,882 and then take the -2,017th root of that (or raise both sides to the power 1/2,017 and invert the lot)...thus:

3000 / 167952,882 = x^-2,017

then

(3000 / 167952,882)^(1 / 2,017) = 1 / x = x^-1

Then just take 1/ answer to get x (I get x = 7.356 to 3 decimal places).

Similarly for 3.

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0 ·Off to shower, remembering forms, and bed.

People wish me luck. I need it. Seriously.

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0 ·- Report

0 ·I just don't get maths. It's like some bizarre foreign language to me. Every time I even think I get close to understanding it, it suddenly makes absolutely no sense.

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0 ·That's all it really is after all: a set of rules and methods you have to learn. Once you learn all the rules and practice the methods for solving problems, you can approach any mathematical problem with the confidence to solve it. Like learning a languge; at first you learn some basic words and phrases to get by and eventually you get to a stage where you correct for things like correct tense and voice. It's similar in math and science. At first you learn over simplified truths but eventually you deal in precision and can handle completely alien problems. That's what I think anyway J and Mr_Wobble.

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0 ·- Report

0 ·I think to be honest that it's about interest. At least that's how it i with me. When i can apply science to something useful I don't have problems. Otherwise it plain bores me and I really can't bother.

When looking at my grades in school it's very noticable what has my interest and what hasn't.

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0 ·nah man your a bit wrong it isnt x-2017 but x^-2017 so you need to do logs of both sides (if u need help with log laws, ask!)

log3000=log(167952,882*x^-2,017)

log3000=log167952,882 + log(x^-2,017)

log3000 - log167952,882 = -2017*log(x)

[ log (3000/167952,882) ] / -2017 = log x

then whatever you get for the left hand side of that, do inverse log, and you have x

and you can use either the natural log (ln) or log(base 10), just use it throughout, and use the same inverse log function!

<doing a stupidly hard maths course for 1st yr uni chemistry with maths, so this sint too much of a challenge, just dont ask me matrices questions!!!!>

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