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
Hi smmmc!
You have to use the logarithm-rules, i.e.
$\displaystyle log(a/b) = log(a) - log(b)$
and $\displaystyle log(a*b) = log(a)+log(b)$
and $\displaystyle log(a^n) = n*log(a)$
Now lets see.
$\displaystyle log(\frac{100x^3y^{-1/2}}{y^2})$
Just using the rules leads to
$\displaystyle log(100x^3y^{-1/2})-log(y^2)$
$\displaystyle log(100x^3y^{-1/2})-2log(y)$
$\displaystyle log(100)+log(x^3)+log(y^{-1/2})-2log(y)$
$\displaystyle log(100)+3log(x)-1/2*log(y) - 2log(y)$
Do you see which rule I used in each step? Did you understand everything?
$\displaystyle log_{10}(100) = ...$ well, you know how to calculate it, right?
Regards
Rapha
You are being asked to apply the rules:
$\displaystyle \log_b \frac{x}{y} = \log_b x - \log_b y$,
$\displaystyle \log_b xy = \log_b x + \log_b y,$
and $\displaystyle \log_b x^k = k\log_b x$.
Apply these rules to the given expression, make the substitutions they give you, and simplify where necessary.
EDIT: Too slow!
I would first rewrite the fraction as a sum/difference of logs:
$\displaystyle \log_{10} \left(\frac{100x^3 y^{-1/2}}{y^2}\right)$
$\displaystyle \begin{aligned}
&= \log_{10} \left(\frac{100x^3}{y^{5/2}}\right) \\
&= \log_{10} (100) + \log_{10} (x^3) - \log_{10} (y^{5/2})
\end{aligned}$
Can you take it from here?
01
EDIT: Ack, beaten to it by three posters!