Two Logarithm Concepts

Ahm97sic SG

There are two logarithm concepts that are not in the usual textbooks, guides or ten years series. However, these two concepts are already known to students and teachers.

Question

(1) Evaluate logn 9 / logn 4

(2) Given that zy = 2^[log2 (z + x)], show that z = x / [y-1]

[The number or letter in red is in subscript ie the base of the log].

These two questions are not dificult as these two questions are just used to illustrate the two logarithm concepts where some students might not be aware.

Thank you for your kind attention.

Regards,

ahm97sic

secretliker SG

(1)

logn 9 / logn 4

= (lg n / lg 9) / (lg n / lg 4)

= (lg n / lg 9) * (lg 4 / lg n)

= lg 4 / lg 9

= lg 2 / lg 3 ?

skythewood SG

try this on calculator... log 9/log 4

now try this.. ln9 / ln 4

one of the first things i teach about logs

Ahm97sic SG

Hi Secretkiller and Skythewood,

The concept used in question 1 is

logn 9 / logn 4 = ratio of the logarithms of any same base ie

logn 9 / logn 4 = log10 9 / log10 4 = loge 9 / loge 4 = log2 9 / log2 4 and so on.

[The number or letter in red is in subscript ie the base of the log].

Thank you for your kind attention.

Regards,

ahm97sic

wishboy SG

zy = 2^[log2 (z + x)]
zy = z + x
zy - z = x
z(y - 1) = x
z = x/(y - 1) (shown)

Ahm97sic SG

Hi Wishboy,

Well done, your answer is perfectly correct.

The concept used in the second question is

a^loga y = y.

So, 2^[log2 (z + x)] = z + x, 3^ (log3 4) = 4, e^ (ln x^2) = x^2, 10^ (lg 7) = 7 and so on

[The number or letter in red is in subscript ie the base of the log].

Thank you for your kind attention.

Regards,

ahm97sic

xiaobai_23 SG

for 1st qns is it tat logn 9 / logn 4=logn9-logn4???

Ahm97sic SG

Hi Xiaobai,

It is logn (9/4) = logn 9 - logn 4. This rule is found in all the textbooks, guides and

ten years series

But, logn 9 / logn 4 is not equal to logn 9 - logn 4.

logn 9 / logn 4 = log10 9 / log10 4 = loge 9 / loge 4 This rule is not found in the

usual textbooks, guides and

ten years series. However,

many students and teachers

have already known about it.

[The number or letter in red is in subscript ie the base of the log].

Thank you for your kind attention.

Regards,

ahm97sic

dadeadman1337 SG

So whats the ans? 1.58?

FirePig SG

(1)

logn 9 / logn 4

= (lg n / lg 9) / (lg n / lg 4)

= (lg n / lg 9) * (lg 4 / lg n)

= lg 4 / lg 9

= lg 2 / lg 3 ?

wrong la.

skythewood SG

So whats the ans? 1.58?

yah, as shown by calculator

eagle SG

thanks ahm97sic

used your question in ExamWorld... under A maths logarithms, with a backlink to this thread and a courtesy thanks to you :D

Ahm97sic SG

Hi Eagle,

You are welcome to include these questions in the examworld.

Thank you for your kind attention.

Regards,

ahm97sic

cantnmousegame SG

err how do i solve these two questions?

(2x)^(lg2) = (7x)^(lg7)

solve (log a X)^(log b X) = x where a, b are positive real numbers except 1. leave ur answers in terms of a and b.

thx

Ahm97sic SG

Hi cantnmousegame,

Question 1

Solve for x in the equation

(2x)^(lg2) = (7x)^(lg7)

Answer

(2x)^(lg2) = (7x)^(lg7)

(2^lg2)(x^lg2)=(7^lg7)(x^lg7)

(x^lg2)/(x^lg7)=(7^lg7)/(2^lg2)

x^(lg2-lg7)=(7^lg7)/(2^lg2)

Bring over the power to the other side of the equation

x = [(7^lg7)/(2^lg2)]^{1/(lg2-lg7)}

Question 2

Solve (log a X)^(log b X) = x where a, b are positive real numbers except 1, leave the answer in terms of a and b.

Answer

(loga x)^(logb x) = x

Let x = b^(logb x)

(loga x)^(logb x) = b^(logb x)

Same power, equate base

loga x = b

x = a^b

The number in red is in subscript ie the base of the log].

The second question is interesting and fun. Please post more in the forum.

Dear cantnmousegame, you should post these two questions in a new thread so that others can have the fun to solve the questions.

Thank your for your kind attention.

Regards,

ahm97sic

proasd SG

continuation of (2x)^(lg2) = (7x)^(lg7)

taking lg on both side,

lg2(lg2+lgx) = lg7(lg7+lgx)
lgx(lg2-lg7)=(lg7)^2-(lg2)^2
lgx(lg2-lg7)=(lg7-lg2)(lg7+lg2)
lgx=-(lg7+lg2)
x=1/14

solved without calc.

LatecomerX SG

There are two logarithm concepts that are not in the usual textbooks, guides or ten years series. However, these two concepts are already known to students and teachers.

Question

(1) Evaluate logn 9 / logn 4

(2) Given that zy = 2^[log2 (z + x)], show that z = x / [y-1]

[The number or letter in red is in subscript ie the base of the log].

These two questions are not dificult as these two questions are just used to illustrate the two logarithm concepts where some students might not be aware.

Thank you for your kind attention.

Regards,

ahm97sic

I'll just solve Qn 1 here since it seems easy to me.

Set n = 4.

log4 9 / log4 4
= log4 9
= 2 log4 3

LatecomerX SG

= log2 3 / 0.5 (log2 4)
= log2 3 / (0.5 * 2)(log2 2)
= log2 3

LatecomerX SG

I just realized the answer to Qn 1 can be summarized into two steps.

logn 9 / logn 4
= 2 logn 3 / 2 logn 2 (coefficients can be cancelled out)
= log2 3 (reversal of "change of base" method)

Ahm97sic SG

Hi LatecomerX,

I agree with your finding ie with the ratio of the logarithm of any same base,

logn 3 / logn 2
= log2 3

Great, you have discovered a logarithm concept that is not found in the usual textbooks, guidebooks and ten years series ! Congratulation !

This forum has not only helped the students to do their homework, it can also help to generate more ideas, new ways and methods, short-cut methods to solve maths questions. Please continue to contribute more to the forum.

Thank you for your kind attention.

Regards,

ahm97sic