Author Topic: Bucket's Questions (Read 56741 times) Tweet Share

dcc · « **Reply #45 on:** February 24, 2008, 11:45:17 pm »

the question must be wrong, because if you manually type the question into the calculator, you get an answer of roughly 10.6 ish

bucket · « **Reply #46 on:** February 24, 2008, 11:48:17 pm »

ah ok, thanks =].

Then how do you work this one out;
$log_4(32) - log_9(27)$

dcc · « **Reply #47 on:** February 24, 2008, 11:49:56 pm »

if the question was $log_2(128) + log_3(45) - log_3(5)$ then the answer would be 9

dcc · « **Reply #48 on:** February 24, 2008, 11:52:01 pm »

Quote from: bucket on February 24, 2008, 11:48:17 pm

$log_4(32) - log_9(27)$

$log_4(2^5) - log_9(3^3)$

Since $2 = 4^\frac{1}{2}$ and $3 = 9^\frac{1}{2}$ , we can rewrite as:

$log_4(4^\frac{5}{2}) - log_9(9^\frac{3}{2})$

$\dfrac{5}{2} - \dfrac{3}{2}$

$= 1$

bucket · « **Reply #49 on:** February 24, 2008, 11:53:49 pm »

Quote from: dcc on February 24, 2008, 11:49:56 pm

if the question was $log_2(128) + log_3(45) - log_3(5)$ then the answer would be 9

That's what I wrote! lol
How is it worked out??

Oh and thanks for solving that other problem.

dcc · « **Reply #50 on:** February 24, 2008, 11:55:41 pm »

Quote from: bucket on February 24, 2008, 11:01:29 pm

Simplify $log_2(128) + log_3(145) - log_3(5)$ .

Ack, I don't know what to do when faced with two different bases lol.

thats what you wrote

bucket · « **Reply #51 on:** February 24, 2008, 11:57:06 pm »

LOL...whoops...sorry :p

dcc · « **Reply #52 on:** February 24, 2008, 11:59:24 pm »

$log_2(128) + log_3(45) - log_3(5)$

we know that $log_2(2^7) = 7$

and from our log laws:

$log_3(45) - log_3(5) = log_3(\dfrac{45}{5}) = log_3(9) = log_3(3^2) = 2$

Adding these, you get 9

bucket · « **Reply #53 on:** February 25, 2008, 01:02:40 am »

Lol thanks for all of that dcc.

bucket · « **Reply #54 on:** February 25, 2008, 08:47:04 pm »

uhm..
Simplify the following expressions:
$(4x^2y^2)^2 + 2(x^2y)^4$
and
$\frac{3(2x^2y^3)^4}{2x^3y^2}$

thanks in advance!

Toothpaste · « **Reply #55 on:** February 25, 2008, 08:53:11 pm »

$(4x^2y^2)^2 + 2(x^2y)^4$

= $2x^8y^4 + 16x^4y^4$

= $2x^4y^4(x^4+8)$

$\frac{3(2x^2y^3)^4}{2x^3y^2}$

= $\frac{3(16x^8y^{12})}{2x^3y^2}$

= $\frac{48x^8y^{12}}{2x^3y^2}$

= $24x^5y^{10}$

bucket · « **Reply #56 on:** February 25, 2008, 10:12:03 pm »

Uhm, can somebody confirm these answers?
Not that I think you're wrong toothpick, it's just that I left my text book at school and the pdf of answers on the cd starts at chapter 14 for some strange reason, and well I don't remember the answers being like that :|
You are probably right though, my memory isn't always reliable.

Collin Li · « **Reply #57 on:** February 25, 2008, 10:25:48 pm »

They're correct, but the first one might not be "simplified." That's sort of subjective. The second line (expanded form) might be considered as simplified. Exam questions aren't this ambiguous - they will specify a "form" to put the answer in (e.g: "Express your answer in the form $a+\sqrt b$ ")

bucket · « **Reply #58 on:** February 25, 2008, 10:26:50 pm »

Okay! Thankyou both!

bucket · « **Reply #59 on:** February 27, 2008, 08:13:17 pm »

Express as the logarithm of a single term:
$1+log_{10}(a)-\frac{1}{3}log_{10}(b)$

edit
also;
$2log_e(x+5)=6$

and
Evaluate:
$log(\sqrt{2})+log_2(1)+2log_2(2)$

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