Author Topic: Bucket's Questions (Read 56855 times) Tweet Share

unknown id · « **Reply #90 on:** March 01, 2008, 07:47:51 pm »

i think you meant "If $log_{10}(x) = a$ and $log_{10}(y) = c$ "

$log_{10}(\frac{100x^3y^{-\frac{1}{2}}}{y^2})$

$= log_{10}(100x^3y^{-\frac{5}{2}})$

$= log_{10}(100) + log_{10}(x^3) + log_{10}(y^{-\frac{5}{2}})$

$= log_{10}(10^2) + log_{10}(x^3) + log_{10}(y^{-\frac{5}{2}})$

$= 2log_{10}(10) + 3log_{10}(x) + {-\frac{5}{2}}log_{10}(y)$

$= 2 + 3a - \frac{5c}{2}$

bucket · « **Reply #91 on:** March 01, 2008, 08:33:46 pm »

Ah shit. I did too.
Sorry man.

Thanks for the help!

bucket · « **Reply #92 on:** March 01, 2008, 11:49:20 pm »

How would you find the intercept and asymptote of the graph $2log_22x+2$

Collin Li · « **Reply #93 on:** March 01, 2008, 11:58:20 pm »

$y = 2\log_22x+2$

$x\mbox{-intercept(s)}$ : $y = 0$

$0 = 2\log_22x+2$

$\implies \log_22x = -1$

$\implies 2x = 2^{-1}$

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

Therefore, the x-intercept is at $\left(\frac{1}{4},\, 0\right)$

$y\mbox{-intercept}$ : $x = 0$

There are no solutions. The domain of this graph is $x > 0$ , so $x=0$ cannot be put into this equation. We know this because we know the logarithm function only accepts values greater than zero.

The asymptote is: $x=0$ . The asymptote for a typical logarithm graph is always $x=0$ unless there have been transformations applied to it (specifically, a translation parallel to the x-axis). In this case, there has been no translation, so the asymptote has not moved. The asymptote corresponds to the "edge" (not mathematical term) of the domain. For example, the domain is $x > 0$ , so the asymptote is $x=0$ .

bucket · « **Reply #94 on:** March 02, 2008, 12:19:47 am »

Cheers coblin.

bucket · « **Reply #95 on:** March 08, 2008, 01:59:03 pm »

$2^{x-1}=3^{x+1}$
solve for x expressing answer as the exact value.

Ahmad · « **Reply #96 on:** March 08, 2008, 02:17:41 pm »

Method 1
Notice that $3 = 2^{\log_2 (3)}$

Therefore, $2^{x-1} = (2^{\log_2 (3)})^{x+1}$

$x - 1 = (x+1) \log_2 {3}$

Method 2
Take log of both sides

$\log 2^{x-1} = \log 3^{x+1}$

$(x-1) \log 2 = (x+1) \log 3$

bucket · « **Reply #97 on:** March 08, 2008, 02:28:46 pm »

thanks ahmed... uhm is there any way to make x the subject?

Glockmeister · « **Reply #98 on:** March 08, 2008, 03:14:14 pm »

With the second method

Divide the Left Hand Side by $(x+1)$ .
Divide the Right Hand Side by $\log 2$

You end up getting $\frac{x-1}{x+1} = \frac{\log 3}{\log 2}$

Then you can use long division to break that up.

_1____
x + 1 | x - 1
x + 1
-------
- 2

Thus $1 - \frac {2}{x+1} = \log_{2} 3 (because \frac{\log 3}{\log 2} = \log_{2} 3 )$

The rest is just simple algebraic manipulation.

Neobeo · « **Reply #99 on:** March 08, 2008, 03:55:30 pm »

Traditionally, you could just expand all the terms and collect the $x$ es to one side.

Glockmeister · « **Reply #100 on:** March 08, 2008, 04:39:06 pm »

True that, what didn't I think of that? lol

Mao · « **Reply #101 on:** March 08, 2008, 08:46:12 pm »

if you multiplied by two on both sides:

$2^x=2 \cdot 3^{x+1}$

then take log₂ on both sides

$x=log_2 2 + log_2 3^{x+1}$

$x=1+(x+1)log_2 3$

$x=log_2 3 \cdot x + 1 + log_2 3$

didnt even touch rational functions

bucket · « **Reply #102 on:** March 08, 2008, 09:39:52 pm »

ah thanks for the help guys!

I have another question =\

If $a log_27 = 3-log_614$ , find the value of a

note:the answer needs to be given in the approximate form, and you can only enter $log_e$ and $log_{10}$ into the calculator.

Toothpaste · « **Reply #103 on:** March 08, 2008, 09:49:08 pm »

change of base law
$alog_2(7) = 3 - (\frac{log_{10}{14}}{log_{10}{6}})$

$\implies alog_2(7) = 1.52711$

$\implies a = \frac{1.52711}{log_2(7)}$

$\implies a = \frac{1.52711}{(\frac{log_{10}{7}}{log_{10}{2}})}$

$\implies \therefore a = 0.543969$

bucket · « **Reply #104 on:** March 08, 2008, 09:51:58 pm »

thanks toothpick you legend.

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