Hey need help on the 3 questions,

thanks heaps

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- Jul 20th 2009, 03:34 AMsmmmcLogarithms
Hey need help on the 3 questions,

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

You have to know that for your basic logarithm function $\displaystyle y = \log_{b}x$ the asymptote is at x = 0. The only way that it would change is if we have a horizontal shift. Do we in $\displaystyle y = 3\log_{2}(5x) + 2$? (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?

01 - Jul 20th 2009, 04:32 AMsmmmc
its a toss up between a and b hrmm

- Jul 20th 2009, 05:03 AMSoroban
Hello, smmmc!

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

Quote:

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

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

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

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

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

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

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

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

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