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**BenLee****Guest**

Hello I have been trying this questions forever to no avail. Could anyone help me?

A. Using the laws of logarithms, simplify x-0.45 = 0.521 and evaluate for x [5 marks]

B. Find the fifth root of 600 using logarithms. [4 marks]

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 103,832

Hi;

A. Using the laws of logarithms, simplify x-0.45 = 0.521 and evaluate for x [5 marks]

Why do you need logarithms to solve a linear equation?

Add .45 to both sides.

For B, what are you allowed to use? A calculator? A log table?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**Benlee****Guest**

Wow... Didn't know it was that simple... haha, but it is an assignment and unfortunately I need to use logarithm for it. I was thinking of putting "lg" on both sides, but not sure if that's the correct way.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 103,832

Do you have the right problem?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**BenLee****Guest**

Apologies, knew something was not right. I typed wrongly.

A. Using the laws of logarithms, simplify** x^-0.45** = 0.521 and evaluate for x [5 marks]

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 103,832

x^-0.45 = 0.521

When you take the log of both sides what do you get?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**BenLee****Guest**

lg x^-0.45 = lg 0.521

Is this right? Seems weird to me.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 103,832

That is correct and what is log(x^(-.45)) ?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**BenLee****Guest**

I got -0.45 lgx = lg 0.521

lg x = 0.629

Is that the answer or do i need to move the lg to make x the subject? o.o

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 103,832

What did you get for lg 0.521?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**BenLee****Guest**

I got -0.283162276 and i divided that by -0.45 to get 0.629

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 103,832

That is correct. So what do you have left?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**BenLee****Guest**

I have lg x = 0.629 left.

Should i make x the subject now or is that the answer? o.o

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 103,832

What is the inverse of log(x)?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**BenLee****Guest**

Is it log x^-1?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 103,832

You are using the common log?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**BenLee****Guest**

I guess so, the other log thingy is "ln"

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 103,832

When we say log(3) we mean what power must 10 be raised to, to get 3.

So what do you think now is the inverse of log? You should get raising 10 to the power of.

So taking 10^ of both sides what do you get?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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