logarithmic notation

**cypress** · October 19th 2011, 10:45 AM

Hi

Could somebody help me solve this question
(a) Express the following statements in logarithmic notation:3.375= 2.25 3/2
That's 3over2 . Thanks so much

**skeeter** · October 20th 2011, 04:32 PM

Originally Posted by cypress

Hi

Could somebody help me solve this question
(a) Express the following statements in logarithmic notation:3.375= 2.25 3/2
That's 3over2 .

$a = b^c$

$c = \log_b{a}$

**cypress** · October 23rd 2011, 11:45 AM

Thanks

But I don't understand how you got that answer...could you break it down?
I would be very greatful.Thanks

**skeeter** · October 23rd 2011, 12:12 PM

Originally Posted by cypress

Thanks

But I don't understand how you got that answer...could you break it down?
I would be very greatful.Thanks

Logarithms: Introduction to "The Relationship"

**cypress** · October 23rd 2011, 12:21 PM

Originally Posted by skeeter

Logarithms: Introduction to "The Relationship"

Thank you so much!!!

**HallsofIvy** · November 4th 2011, 06:36 AM

cypress, what skeeter wrote was not an "answer" but the basic relationship between a logarithm and an exponential. It is impossible to give an answer to the problem you posted because you clearly did not post it correctly- there is no exponent in what you gave so it can't be converted to logarithms. Did you mean to write 3.375= 2.25^(3/2)?

**cypress** · November 26th 2011, 02:53 AM

Originally Posted by HallsofIvy

cypress, what skeeter wrote was not an "answer" but the basic relationship between a logarithm and an exponential. It is impossible to give an answer to the problem you posted because you clearly did not post it correctly- there is no exponent in what you gave so it can't be converted to logarithms. Did you mean to write 3.375= 2.25^(3/2)?

Yes thats what i was trying to workout. Wolud you mind showing me how it's done?
Thanks

**Siron** · November 26th 2011, 03:55 AM

Take the logarithm of both sides, therefore:
$\log(3,375)=\log(2,25^{\frac{3}{2}})$
Now use the following rules to continue:
$\log(a^b)=b\log(a)$
$\frac{\log(b)}{\log(a)}=\log_a(b)$

**cypress** · November 27th 2011, 03:52 AM

Originally Posted by Siron

Take the logarithm of both sides, therefore:
$\log(3,375)=\log(2,25^{\frac{3}{2}})$
Now use the following rules to continue:
$\log(a^b)=b\log(a)$
$\frac{\log(b)}{\log(a)}=\log_a(b)$

Hi

I'm new to logarithms,unsure how to do the next step. Can you show me or do you know a link I could learn from?

Thanks so much

**Siron** · November 27th 2011, 04:33 AM

If you have to solve this exercice then I thought you would know some rules about logarithms. But have you tried something with the rules I mentioned?
$\log(3,375)=\log(2,25^{\frac{3}{2}})$
$\Leftrightarrow \log(3,375)=\frac{3}{2}\log(2,25)$ (is this clear?)
$\Leftrightarrow \frac{\log(3,375)}{\log(2,25)}=\frac{3}{2}$

Now, apply the last rule I mentioned.

**cypress** · November 27th 2011, 05:15 AM

Tha much is clear, can't grasp the concept of the next step :-(

**skeeter** · November 27th 2011, 05:38 AM

in your original post, you started with an exponential equation ...

$3.375 = 2.25^{3/2}$

... and were asked to change it to an equation using logarithmic notation

$\log_{2.25}(3.375) = \frac{3}{2}$

that's all.

**cypress** · November 27th 2011, 05:42 AM

I can understand it when I see it broken down, I find it difficult to know what rules to use.

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