0
$\begingroup$

I'm still having some trouble fully understanding log expressions.

$$T(n) = c \times n^2 log_{10} n $$

I want to rearrange to find the value of $c$, when $n = 100$ and $T(n) = 10$ but not entirely sure of how to rearrange the expression.

This is what I have tried: $$c = \frac{T(n) \times n} {\log_{10} \times {n^2}} $$

Thanks in advance.

$\endgroup$
3
$\begingroup$

$\log_{10}$ is not a number on its own, so you cannot multiply/divide/add/subtract by it. Consider it as a function like Sin(x). Sin on its own doesn't make any sense, and neither does log. They require a number to be operated on, such as $\log_{10}100$ or $\sin{(\pi/2)}$

Consider your equation:

$$T = c \times n^2 log_{10} n$$ $$\implies c = \dfrac{T}{n^2 log_{10} n}$$ In the above I just divided both sides by the "junk" multiplied with $c$.

Now since $n=100$, $\log_{10}{n}=\log_{10}{100}$, which can be interpreted like this:

What is $x$ such that $10^x=100$?

In general for $\log_{a}{b}$ means: find $x$ such that $a^x=b$

The answer for "What is $x$ such that $10^x=100$?" is obviously $2$.

$$\therefore \log_{10}{100}=2$$

Now, again, consider:

$$c = \dfrac{T}{n^2 log_{10} n}$$ $$\implies c=\dfrac{10}{100^2\cdot 2}=\dfrac{1}{2000}$$

$\endgroup$
0
0
$\begingroup$

take $\log_{10} n = y$

$ T = c * n^2 * y$

$c = \cfrac {T}{n^2 * y}$

Now substitute y here,

$c = \cfrac {T}{n^2 \log_{10} n}$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.