O-Level Additional Math

[iFruit]

SKT:

Hi,


Solve the equation
lg3 + 2lg2x = lg(x + 6)

TIA

lg3 + 2lg2x = lg(x+6)

lg3 + lg(2x)² = lg(x+6)

lg3.(2x)² =lg(x+6)

12x² = x+6

12x² -x -6 = 0

(4x-3)(3x+2) = 0

x = 3/4 or -2/3

[SKT]

iFruit:

SKT:

Hi,


Solve the equation
lg3 + 2lg2x = lg(x + 6)

TIA

lg3 + 2lg2x = lg(x+6)

lg3 + lg(2x)² = lg(x+6)

lg3.(2x)² =lg(x+6)

12x² = x+6

12x² -x -6 = 0

(4x-3)(3x+2) = 0

x = 3/4 or -2/3

Hi,

The answer provided is x = 3/4 only. Could it be x = -2/3 is invalid?

TIA

[iFruit]

SKT:

iFruit:

[quote=\"SKT\"]Hi,


Solve the equation
lg3 + 2lg2x = lg(x + 6)

TIA

lg3 + 2lg2x = lg(x+6)

lg3 + lg(2x)² = lg(x+6)

lg3.(2x)² =lg(x+6)

12x² = x+6

12x² -x -6 = 0

(4x-3)(3x+2) = 0

x = 3/4 or -2/3

Hi,

The answer provided is x = 3/4 only. Could it be x = -2/3 is invalid?

TIA[/quote]Yes, -2/3 is invalid as log of a negative number is undefined. Sorry for the oversight.

[SKT]

iFruit:

SKT:

[quote=\"iFruit\"]
lg3 + 2lg2x = lg(x+6)

lg3 + lg(2x)² = lg(x+6)

lg3.(2x)² =lg(x+6)

12x² = x+6

12x² -x -6 = 0

(4x-3)(3x+2) = 0

x = 3/4 or -2/3

Hi,

The answer provided is x = 3/4 only. Could it be x = -2/3 is invalid?

TIA

Yes, -2/3 is invalid as log of a negative number is undefined. Sorry for the oversight.[/quote]But 2lg2x = lg(2x)², (2x)² ≥ 0 ?

TIA

[iFruit]

SKT:

iFruit:

[quote=\"SKT\"]
Hi,

The answer provided is x = 3/4 only. Could it be x = -2/3 is invalid?

TIA

Yes, -2/3 is invalid as log of a negative number is undefined. Sorry for the oversight.

But 2lg2x = lg(2x)², (2x)² ≥ 0 ?

TIA[/quote]sure, (2x)² ≥ 0 but when lg(x) is used, we can use only the positive values of x.

so when x=-2/3, lg(x) is undefined->lg(2x) is undefined->2lg(2x) is undefined, even though lg(16/9) is defined.

lg(16/9) can only be 2log(4/3) and not 2log(-4/3)

HTH

[hot_chocolate]

Please help to solve the followings:


1) The product of n whole numbers 1 x 2 x 3 x 4 x 5 x …x (n - 1) x n has 28 consecutive zeros. Find the largest value of n.

2) Find the largest no. n such that there is only one whole no. k that satisfies
8/21 < n/(n+k) < 5/13

(Note: A < C < B means that value of C is between A and B, e.g. 4 < 9 < 16)

Thank you.

[atutor2001]

hot_chocolate:

Please help to solve the followings:


1) The product of n whole numbers 1 x 2 x 3 x 4 x 5 x ....x (n - 1) x n has 28 consecutive zeros. Find the largest value of n.

Using explanation from this link
http://2000clicks.com/MathHelp/BasicFactorialConsecutiveIntegerProducts.aspx

You may like to read it and let me know your answer. I tried but can't get consecutive zeros

[iFruit]

hot_chocolate:

Please help to solve the followings:


1) The product of n whole numbers 1 x 2 x 3 x 4 x 5 x ....x (n - 1) x n has 28 consecutive zeros. Find the largest value of n.

In a factorial, every multiple of 5 will contribute one zero at the end

for example, 1x2x3x4x5 = 120, 120x6x7x8x9x10 = [something]00

In addition every multiple of 25 will contribute one extra zero

25x24, 50x48, 75 x 72, 100x99 etc.

So 120! would have 120/5 = 24 zeros contributed by multiples of 5,

and 4 extra zeros contributed by multiples of 25 (25, 50, 75, 100) with 28 zeros (24+4) at the end.

So the largest factorial with 28 consecutive zeros is 124! --> n = 124

[iFruit]

hot_chocolate:

Please help to solve the followings:



2) Find the largest no. n such that there is only one whole no. k that satisfies
8/21 < n/(n+k) < 5/13

(Note: A < C < B means that value of C is between A and B, e.g. 4 < 9 < 16)

Thank you.

8/21 < n/(n+k) < 5/13 ---> 8/21 < 1/(n+k)/n < 5/13 ---->

8/21 < 1/(1+k/n) < 5/13 ---> 1/(21/8 ) < 1/(1+k/n) < 1/(13/5) ---->

13/5 < 1 + k/n < 21/8 ----> 8/5 < k/n < 13/8 ----> 64/40 < k/n < 65/40


Now, 64/40 < k/n < 65/40 has no solutions for k = whole number, n=40

128/80 < k/n < 130/80, will have one solution for k=129, n =80

192/120 < k/n < 195/120 will have multiple solutions for k (193, 194), and n =120

so \" largest n such that there is only one whole no. k \" = 80

HTH

[atutor2001]

iFruit:

hot_chocolate:

Please help to solve the followings:


1) The product of n whole numbers 1 x 2 x 3 x 4 x 5 x ....x (n - 1) x n has 28 consecutive zeros. Find the largest value of n.

In a factorial, every multiple of 5 will contribute one zero at the end

for example, 1x2x3x4x5 = 120, 120x6x7x8x9x10 = [something]00

In addition every multiple of 25 will contribute one extra zero

25x24, 50x48, 75 x 72, 100x99 etc.

So 120! would have 120/5 = 24 zeros contributed by multiples of 5,

and 4 extra zeros contributed by multiples of 25 (25, 50, 75, 100) with 28 zeros (24+4) at the end.

So the largest factorial with 28 consecutive zeros is 124! --> n = 124

Hi iFruit

You are really good!