indices and logarithms

[Vastor]

hey guys,

 

well, as we know, adding among log like this is easy

 

[math]\log_c A + \log_c B = \log_c AB[/math]

 

but how does to "add" something like this?

 

[math]\log_3 12-3^x = x - 1[/math]

 

and basically, how you calculate add in an equation that has indices and logarithms?

 

I mean if

 

[math]A * A = A^{1+1}[/math]

 

but how about

[math]A + A = A^? = 2A[/math]

[ajb]

Equalities like

 

[math]x = \log_{a}(x^{a}) = a^{\log_{a}(x)}[/math],

 

maybe useful.

 

______________________

 

There us an obvious typo here!

 

[math]x = \log_{a}(a^{x})[/math]

Edited February 15, 2012 by ajb

[Fuzzwood]

Another hint: 12 = 9 + 3

[Vastor]

ouch, I don't know what the cause, my brain hurt so hard when dealing with logs that involve adding like this question, indices is much better alternative

 

[math]3^{x-1} + 3^x = 12 [/math](originals of above question)

 

using factorisation and other stuff without involving log can be solved easily and get the answer of x = 2.

 

but can log solve above problem also? I try but I had problem when I'm trying to convert the indices to a log because of the add/minus symbol. it give me so much headache on how to create a symbol for the add/minus (like multiply and divide they turn into plus/minus in log) if then, what plus/minus turn out to be?

 

even using ajb recommendation, I still can't convert it in which I'm sure I'm do it right

 

if [math] x = \log_{a}(x^{a}) = a^{\log_{a}(x)} [/math]

 

then [math] 3^{x-1} + 3^x = \log_{1}{12^1}[/math]

i'm doing it right? if then, how I convert on the left part to log? arrrghh

[Vastor]

anyone?

[ajb]

Okay. A good hint would be to multiply both sides of your equation by [math]3^{1-x}[/math]. Use the law of indices and then take your logs.

[Vastor]

Okay. A good hint would be to multiply both sides of your equation by [math]3^{1-x}[/math]. Use the law of indices and then take your logs.

ok, I got it, thnx, so the clue here is to put the variable x in one place only(aka not seperated between two number by the plus symbol)?!

[ajb]

I think so.

 

One is tempted to write

 

[math]\log(a+b) = \log(a)+ log(b)[/math],

 

but this is not true. I cannot see any more direct way of using logs in tackling this question. Though note everything is quite easy in your example and the answer can just be spotted.

[Vastor]

ok, I got it, thnx, so the clue here is to put the variable x in one place only(aka not seperated between two number by the plus symbol)?!

forget this, this one is clue on how to solve the question's problem...

 

I think so.

 

One is tempted to write

 

[math]\log(a+b) = \log(a)+ log(b)[/math],

 

but this is not true. I cannot see any more direct way of using logs in tackling this question. Though note everything is quite easy in your example and the answer can just be spotted.

 

I understand that, when first seeing log, I can't really see the pattern on how it work, and can't even understand why

 

[math]A^B = C [/math]

become

[math] log_A C = B[/math]

 

until now, the only thing not really clear to me is what does A represent (which why give me so much difficulty on converting from indice to log or otherwise), yes it is "base" but what is it actually, does it refer to the same "base" of "number type"(sorry, not really know many definition on math here), I mean does it refer to base-ten / base-two number?

 

if so, why does

[math]log_1 12^1 = 12[/math]

does not 12 is number base-10? why it is 1 in log?

 

for the part that I understand on log:-

- they behave like indices, except they are being as the "main number". example:-

[math]10^5 = 100,000[/math]

become

[math]log 100,000 = 5[/math] the 5 became the "main number", and the "original main number" become inner-side of the log.

 

- we can change the number base(and the "original main number") as we like without changing its "main number" values. examples:-

[math]1 = log 10 = log_5 5 = log_6 6 = 1[/math]

 

- there some pattern on how there some relation between the "base" and "main number" and "inner-side number"(all number, nuff said)in log

[math]log_9 \frac{x}{y}[/math]

 

[math]=\frac{log_3 \frac{x}{y}}{log_3 9}[/math]

 

[math]=\frac{log_3 \frac{x}{y}}{2}[/math]

 

[math]=\frac{log_3 x - log_3 y}{2}[/math]

 

[math]=\frac{1}{2}(log_3 x - log_3 y)[/math]

 

[math]=log_3 \sqrt{x} - log_3 \sqrt{y}[/math]

 

can you say something about this?(wikipedia just not enough)

 

edit: should proof-read first

Edited February 15, 2012 by Vastor

[ajb]

There is a typo earlier on, should have spotted that before.

 

The definition of of the log is

 

[math]x= \log_{a}(a^{x})[/math].

 

Now, log to the base one is not defined. One to any power is still one.

[Vastor]

edit : Ignore this post, should have read question better

 

hey guys,

 

I don't know if the problem is in the answer provided in the book but the exercise said

 

Solve the following equation [math] 9log_3 x = 7 [/math]

 

[math] 9log_3 x = 7 [/math]

[math] log_3 x = \frac{7}{9} [/math]

 

[math] 3^{\frac{7}{9}} = x [/math]

 

answer in the book

[math] x = \sqrt{7}[/math]

 

ooo wait, does it legitimate to use graphing calculator like this to find the answer?

lost my calculator somewhere

Edited February 17, 2012 by Vastor

[ajb]

Your answer [math]x= 3^{\frac{7}{9}}[/math] is correct.

[Vastor]

 

how can this happened?

[Vastor]

hey guys,

ignore my post before and never use your pc calculator to calculate log xD

 

anyway, I got another sort of hard question:

 

Solve the equation [math] 4^x - 7*2^x - 8 = 0[/math]

 

[math] 4^x - 7*2^x - 8 = 0 [/math]

 

[math] 4^x - 7*2^x = 8 [/math]

 

[math] 2^x(2^x - 7) = 2^3 [/math]

 

[math] 2^x - 7 = \frac{2^3}{2^x} [/math]

 

using trial and error, you can get x = 3

 

but how to calculate it?

Edited February 19, 2012 by Vastor

[Fuzzwood]

You could pull 7 apart in 2³ - 2⁰. Then you have everything as 2 to the power of something, and something to the power of 0 which always equals 1:

 

x - (3 - 1) = 3/x

 

x - 3 + 1 = 3/x

 

x - 2 = 3/x

 

x² - 2x - 3 = 0

 

(x - 3)(x + 1)

 

x = 3 or x = -1

[Vastor]

You could pull 7 apart in 2³ - 2⁰. Then you have everything as 2 to the power of something, and something to the power of 0 which always equals 1:

 

x - (3 - 1) = 3/x

 

x - 3 + 1 = 3/x

 

x - 2 = 3/x

 

x² - 2x - 3 = 0

 

(x - 3)(x + 1)

 

x = 3 or x = -1

 

don't you think there error in your calculation?

 

how do you get

 

x - (3-1) = 3/x ?

[imatfaal]

To check answers when unsure always sub back into the original equation

 

[math] 4^x - 7*2^x - 8 = 0 [/math]

 

[math] 4^3 - 7*2^3 - 8 = 0 [/math]

 

[math] 64 - 7*8 - 8 = 0 [/math]

 

[math] 64 - 56 - 8 = 0 [/math]

 

ok that's cool

 

[math] 4^{-1} - 7*2^{-1} - 8 = 0 [/math]

 

[math] 1/4 - 7*(1/2) - 8 = 0 [/math]

 

[math]1/4 - 7/2 - 8 = 0 [/math]

 

and that is not right!

 

the error in Fswd's answer is in the taking of logs of multiplied factors

 

I would do it this way

 

[math] 4^x - 7*2^x - 8 = 0 [/math]

 

[math] 2^{2x} - 2^x(2^3-2^0) - 2^3 = 0 [/math]

 

[math] 2^{2x} - 2^x.2^3+2^x.2^0 - 2^3 = 0 [/math]

 

and bracketed to make the correct removal of the 2s clearer

 

[math] 2^{2x} - (2^x.2^3)+(2^x.2^0) - 2^3 = 0 [/math]

 

[math] 2x - (x+3)+(x+0) - 3 = 0 [/math]

 

[math] 2x - x-3+x - 3 = 0 [/math]

 

[math] 2x -3- 3 = 0 [/math]

 

[math] 2x = 6 [/math]

 

[math] x=3 [/math]

[Vastor]

Hey guys,

35. It is known that x and y are related by the equation [math] y = ax^n [/math], where a and n are constants. When a graph of [math] log_2 y [/math] against [math] log_2 x [/math] is plotted, a straight line passing through the points (1,5) and (3,11) is obtained. Find the values of a and n.

 

[math] y = ax^n \Rightarrow lg y = lg a + n*lg x \Rightarrow log_2 y = n*log_2 x + log_2 a[/math], right?

 

[math] log_2 y = m*log_2 x + C[/math], P_1(3, 11), P_2(1, 5)

 

[math] m = \frac{11-5}{3-1} = \frac{6}{2} = 3 [/math] (real answer = 3)

 

if x = 1, y = 5.

[math] log_2 5 = 3*log_2 1 + C \Rightarrow lg 5 = lg 1 + C \Rightarrow lg 5 = C [/math]

 

where C = lg a

[math] lg 5 = lg a \Rightarrow a = 5 [/math] (real answer = 4)

 

 

if x = 3, y = 11.

[math] log_2 11 = 3*log_2 3 + C \Rightarrow lg 11 = 3*lg 3 + C \Rightarrow C = lg 11 - lg 27 \neq lg 5 [/math] how this happen? does Linear Law breaks down its "accurate" relation? or just my daily 'mathypo'.