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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: 90,866

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.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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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: 90,866

Do you have the right problem?

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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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: 90,866

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.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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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: 90,866

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

**In mathematics, you don't understand things. You just get used to them.**

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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: 90,866

What did you get for lg 0.521?

**In mathematics, you don't understand things. You just get used to them.**

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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: 90,866

That is correct. So what do you have left?

**In mathematics, you don't understand things. You just get used to them.**

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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: 90,866

What is the inverse of log(x)?

**In mathematics, you don't understand things. You just get used to them.**

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**BenLee****Guest**

Is it log x^-1?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,866

You are using the common log?

**In mathematics, you don't understand things. You just get used to them.**

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**BenLee****Guest**

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

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,866

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.**

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