ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Halil on February 13, 2011, 08:15:36 pm

Title: Halils Methods questions
Post by: Halil on February 13, 2011, 08:15:36 pm: Just like i do for specialist ill post up questions here that I require help with.

Start with this one:

((log(x,10))/(log(2,10)))=4 find x

answer is 16, its a few clicks answer with the calculator, but what is the manual solution?
Title: Re: Halils Methods questions
Post by: simonhu81292 on February 13, 2011, 08:42:40 pm: isn't the answer 2^(1/4)?
Title: Re: Halils Methods questions
Post by: xZero on February 13, 2011, 08:54:02 pm: This requires the knowledge of base changing in logs (no longer required since CAS has the ability to change base :P )
$\frac{\log_{x} 10}{\log_{2} 10}=4$

$\log_{x} 10=4\log_{2}10$

$\frac{\log_{e} 10}{\log_{e} x}=\frac{4\log_{e} 10}{\log_{e} 2}$

$\frac{1}{\log_{e} x}=\frac{4}{log_{e}{2}}$

$\log_{e} 2=4\log_{e} x$

$\log_{e} x^4=\log_{e} 2$

$x^4=2$

$x=2^{\frac{1}{4}}$
Title: Re: Halils Methods questions
Post by: Halil on February 13, 2011, 09:01:14 pm: no, switch the positions of the x and 10, and the 2 and 10.
lol
Title: Re: Halils Methods questions
Post by: m@tty on February 13, 2011, 09:06:17 pm: I would go:

$\frac{\log_x10}{\log_210}=4$

$\implies \frac{\log_{10}2}{\log_{10}x}=\frac{1}{4}$ (reciprocate both sides) - see note below

$\implies \log_x2=\frac{1}{4}$

$x=2^{\frac{1}{4}}$
It's a bit shorter than your method, xZero :P

Note:

When "flipping" a log, the base and number inside the log are switched:

ie. $\left(\log_ab\right)^{-1}=\frac{1}{\log_ba}$

So when the log fraction is flipped we switch the base and "number" in each.
Title: Re: Halils Methods questions
Post by: simonhu81292 on February 13, 2011, 09:07:12 pm: you would approach it the same was as shown
log10^x = 4 x log10^2
same base..log 10 cancels out
x = 2^4
x= 16
Title: Re: Halils Methods questions
Post by: m@tty on February 13, 2011, 09:08:25 pm: Quote from: Halil on February 13, 2011, 09:01:14 pm
no, switch the positions of the x and 10, and the 2 and 10.
lol

So the question is meant to be

$\frac{\log_{10}x}{\log_{10}2}=4$ ???

If so, then use the change of base rule

$\implies \log_2x=4$

$x=2^4=16$
Title: Re: Halils Methods questions
Post by: Halil on February 13, 2011, 10:33:49 pm: thanks for that M@tty, and others who tried :)
Title: Re: Halils Methods questions
Post by: Halil on February 14, 2011, 11:05:39 pm: Next questions are related with the Log chapter.

Couldn't figure our how to do these two:

solve $4^{2x-b}=20$ for x, where b is subset (the e looking icon thats facing right) of R - $b \in \mathbb{R}$

and solve $2^{x-1}=3^{x+a}$ for x, where a is subset of R - $a \in \mathbb{R}$

Ill try hard to find the solution, but after doing a 5.5 hour study on methods, my brain just stopped.

Any help will be appreciated greatly

(btw, it is 4 to the power of 2x-b, and 2 to the power of x-1 and 3 to the power of x+a, i dont know why the thing made it look like that)

Mod Edit: Fixed LaTeX

PS. Try curly brackets {...} when the power has multiple characters (or log base etc. $\log_10x \text{ vs. } \log_{10}x$ )
Title: Re: Halils Methods questions
Post by: m@tty on February 14, 2011, 11:20:49 pm: 1.

$4^{2x-b}=20$

$\implies 2x-b=\log_4(20)$

$\implies x=\frac{1}{2}\left(\log_4(20)+b\right)$
Title: Re: Halils Methods questions
Post by: kamil9876 on February 14, 2011, 11:24:17 pm: $4^{2x-b}=20$ means that

$2x-b=log_4 20$

$<br />x=(log_4 20 + b)/2$

2)

$2^{x-1}=3^{x+a}$

$log_{10} 2^{x-1}=log_{10} 3^{x+a}$

$(x-1)log_{10} 2=(x+a) log_{10} 3$

Now solve for $x$

edit: you could've taken some other base, 10 was arbitrary, it's just a habit developed from using shit calculators.
Title: Re: Halils Methods questions
Post by: Halil on February 14, 2011, 11:25:36 pm: The answer for that first one is apparantly $((loge(20)+2*bloge(2))/(4loge(2)))$

:S
Title: Re: Halils Methods questions
Post by: m@tty on February 14, 2011, 11:26:22 pm: $2^{x-1}=3^{x+a}$

$\implies \log_2\left(2^{x-1}\right)=\log_2\left(3^{x+a}\right)$

$\implies x-1=(x+a)\log_23$

$x(\log_23-1)=-(1+a\log_23)$

$x=-\frac{1+a\log_23}{\log_23-1}$

$x=\frac{1+a\log_23}{1-\log_23}$