# Thread: Find in terms of a and c for logs

1. ## Find in terms of a and c for logs

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

2. 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

3. 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!

4. 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?

5. 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}
&= \log_{10} \left(\frac{100x^3}{y^{5/2}}\right) \\
&= \log_{10} (100) + \log_{10} (x^3) - \log_{10} (y^{5/2})
\end{aligned}

Can you take it from here?

01

EDIT: Ack, beaten to it by three posters!

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