# Logarithms

• Jul 20th 2009, 03:34 AM
smmmc
Logarithms
Hey need help on the 3 questions,

thanks heaps
• Jul 20th 2009, 03:47 AM
yeongil
$y = 3\log_{2}(5x) + 2$

You have to know that for your basic logarithm function $y = \log_{b}x$ the asymptote is at x = 0. The only way that it would change is if we have a horizontal shift. Do we in $y = 3\log_{2}(5x) + 2$? (In other words what do the 3, 5, and +2 do?) If you can answer that, then you have the answer to the problem.

As for #9,
Quote:

Which one of the following functions has a graph with a veritcal asymptote with equation x = b?
Here's a hint: two of the choices can be eliminated because they do not have vertical asymptotes at all (instead, they only have a horizontal asymptote). Do you know which 2?

01
• Jul 20th 2009, 04:32 AM
smmmc
its a toss up between a and b hrmm
• Jul 20th 2009, 05:03 AM
Soroban
Hello, smmmc!

I have a solution for #8, but there must be a better way.

Quote:

Given $3^x \:=\:4^y \:=\:12^z$, show that: . $z \:=\:\frac{xy}{x+y}$

We have: . $3^x \:=\:12^z$

. . Take logs: . $\log(3^x) \:=\:\log(12^z) \quad\Rightarrow\quad x\log3 \:=\:z\log12$

. . And we have: . $\log3 \:=\:\frac{z\log12}{x}$ .[1]

We have: . $4^y \:=\:12^z$

. . Take logs: . $\log(4^y) \:=\:\log(12^z) \quad\Rightarrow\quad y\log4 \:=\:z\log12$

. . And we have: . $\log4 \:=\:\frac{z\log12}{y}$ .[2]

$\text{Add }{\color{blue}[1]}\text{ and }{\color{blue}[2]}\!:\quad \underbrace{\log3 + \log 4}_{\text{This is }\log12} \;=\;\frac{z\log12}{x} + \frac{z\log12}{y} \;=\;z\log12\left(\frac{x+y}{xy}\right)$

Hence: . $\log12 \;=\;z\log12\left(\frac{x+y}{xy}\right) \quad\Rightarrow\quad 1 \;=\;z\left(\frac{x+y}{xy}\right)$

. . Therefore: . $z \;=\;\frac{xy}{x+y}$