A Math questions that has bugged me for a while

AmuletxHeart SG

How to do question 7, 8c and 9b?

Answers are very much appreciated.

AmuletxHeart SG

ChoCoChips SG

hi.

lets look at qn 7.

first u need to find out the value of c.

so what u do here is u put in values of x and y. (which in this case is 2 and 3, respectively.)

therefore

log2 4 = 2 log2 2 + c

we know 2 log2 2 = log2 4.

so c = 0.

so what's left is

log2 (y+1) = 2 log2 (x)

therefore

y+1 = x^2

y = x^2 - 1.

dunno correct anot.

heh

ChoCoChips SG

qn 9b.

(log5 4)(log2 10)

----------------------

(log25 10^0.5)

here we have 2 logs of different bases.

so we need to change them.

in this case we change all of them to log25.

log5 4 is therefore log25 4/log25 5

log2 10 is therefore log25 2/log25 10

so we have

(log25 4/log25 5) X (log25 10/log25 2)

---------------------------------------------------------

(log25 10^0.5)

log25 4 = 2 log25 2

log25 5 = 0.5 --------------> cuz 5 is the sqrt of 25.

log25 10^0.5 = 0.5 log25 10

with these in mind, we can now simplify the equation.

eventually we will arrive at

(2/0.5) X (log25 10)

------------------------------

0.5 log25 10

which gives us

4 log25 10

-----------------

0.5 log25 10

= 8.

ChoCoChips SG

qn 8c.

(xe^x)^2 = 30e^-x

expanding the LHS,

we get (x^2)(e^2x)

now lets remove the e-x on the RHS

so the equation now is

(x^2)(e^3x) = 30

ln(x^2)(e^3x) = ln 30

then we can separate the LHS, as ln(ab) = ln a + ln b.

ln (x^2) + ln (e^3x) = ln 30

2 ln x + 3x = ln 30

ln x = 3x/2 + (ln30)/2.

hope i ish correct.

heh

wee_ws SG

Hi AmuletxHeart,

It will also be good to reflect on what learning gaps you have, after learning about the solutions. Thanks.

Cheers,
Wen Shih

Hitman+ SG

chocochips.... u mean u still remeber natural log and normal log after so many decades?

Hitman+ SG

remembers macross saga series only ..................throws fat white a maths book away!

quailmaster SG

check out this webpage for some notes on logarithms

http://www.a-maths-tuition.com/2010/05/logarithms-part-iii.html

AmuletxHeart SG

I'm sorry for replying late, but today is the last day of school so I finally have time to revisit this forum. I will do a reflection on this topic ASAP, including areas which I'm always stuck at.