Find in terms of a and c for logs

**smmmc** · July 18th 2009, 08:04 PM

Hey

I dont understand what the question is asking, do i sub the fraction into the log10(x)=a, the x part... ?
above attached

thanks alot

**Rapha** · July 18th 2009, 08:12 PM

Hi smmmc!

Originally Posted by smmmc

Hey

I dont understand what the question is asking, do i sub the fraction into the log10(x)=a, the x part... ?
above attached

thanks alot

You have to use the logarithm-rules, i.e.

$log(a/b) = log(a) - log(b)$

and $log(a*b) = log(a)+log(b)$

and $log(a^n) = n*log(a)$

Now lets see.

$log(\frac{100x^3y^{-1/2}}{y^2})$

Just using the rules leads to

$log(100x^3y^{-1/2})-log(y^2)$

$log(100x^3y^{-1/2})-2log(y)$

$log(100)+log(x^3)+log(y^{-1/2})-2log(y)$

$log(100)+3log(x)-1/2*log(y) - 2log(y)$

Do you see which rule I used in each step? Did you understand everything?

$log_{10}(100) = ...$ well, you know how to calculate it, right?

Regards
Rapha

**AlephZero** · July 18th 2009, 08:13 PM

You are being asked to apply the rules:

$\log_b \frac{x}{y} = \log_b x - \log_b y$ ,

$\log_b xy = \log_b x + \log_b y,$

and $\log_b x^k = k\log_b x$ .

Apply these rules to the given expression, make the substitutions they give you, and simplify where necessary.

EDIT: Too slow!

**VonNemo19** · July 18th 2009, 08:13 PM

Originally Posted by smmmc

Hey

I dont understand what the question is asking, do i sub the fraction into the log10(x)=a, the x part... ?
above attached

thanks alot

$=(\log_{10}100x^3y^{1/2})-(\log_{10}y^2)=[(\log_{10}100)+(\log_{10}x^3)+(\log_{10}y^{-1/2})]+\log_{10}y^2$

Bring down all the exponents and........ can you substitute and simplify?

**yeongil** · July 18th 2009, 08:14 PM

I would first rewrite the fraction as a sum/difference of logs:
$\log_{10} \left(\frac{100x^3 y^{-1/2}}{y^2}\right)$
$\begin{aligned}<br /> &= \log_{10} \left(\frac{100x^3}{y^{5/2}}\right) \\<br /> &= \log_{10} (100) + \log_{10} (x^3) - \log_{10} (y^{5/2})<br /> \end{aligned}$
Can you take it from here?

01

EDIT: Ack, beaten to it by three posters!

**smmmc** · July 18th 2009, 08:29 PM

Hey thank you all ! Thanks for providing me hints and directing me to the right way

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