# Thread: [Functions and Relations - logarithm

1. ## [Functions and Relations - logarithm

Alright, Grade 11b High school math...

bring home assignment (fun stuff)
I’m having a lot of problems with this sheet, hopefully you guys will help me understand the problems. (assignment value is 60 counting on test marks.. due tomorrow which is Monday...)

I’d like to know if my math is correct on this question...

3^(2x-5) = 15^(2-x)
log 3^(2x-5) = log 15^(2-x)
(2x-5)*log 3 = (2-x)*log 15
2x*log 3 - 5*log 3 = 2*log 15 - x*log 15

2x/5 = 2/x (cross multi)

3x/3 = 10/3 = 3.3333 ....

also i have no idea how to tackle this question...

2/3 [12*logx a - 6(logx (4 squar root of:[b])-2*logx c^3)]

i have more, but this is the most puzzling to me...

2. Originally Posted by jessyc
Alright, Grade 11b High school math...

bring home assignment (fun stuff)
I’m having a lot of problems with this sheet, hopefully you guys will help me understand the problems. (assignment value is 60 counting on test marks.. due tomorrow which is Monday...)

I’d like to know if my math is correct on this question...

3^(2x-5) = 15^(2-x)
log 3^(2x-5) = log 15^(2-x)
(2x-5)*log 3 = (2-x)*log 15
2x*log 3 - 5*log 3 = 2*log 15 - x*log 15

2x/5 = 2/x (cross multi)

3x/3 = 10/3 = 3.3333 ....
you were good up to the third to last line, after that, i don't knw what happened. how did you even cross-multiply there?

Picking up where you left off. i will do some rounding, so you may want to redo the steps using more decimal places if you want a more accurate answer

$2x \log 3 - 5 \log 3 = 2 \log 15 - x \log 15$ ....put all the x's on one side

$\Rightarrow 2x \log 3 + x \log 15 = 5 \log 3 + 2 \log 15$ ....work out the logs with your calculator

$\Rightarrow 0.9542x + 1.17609x = 2.385606 + 2.352183$

$\Rightarrow 2.13029x = 4.7378$

$\Rightarrow x = \frac {4.7378}{2.13029}$

$\Rightarrow x = 2.224$

also i have no idea how to tackle this question...

2/3 [12*logx a - 6(logx (4 squar root of:[b])-2*logx c^3)]
Not sure what is going on here either.

do you mean $\frac {2}{3} \left[ 12 \log_x a - 6 \log_x \sqrt [4] {b} - 2 \log_x c^3 \right]$ ?

if so, we need more info. what are a,b, and c? and what exactly do you want to find?

3. Sorry for
2/3 [12*logx a - 6(logx (4 squar root of:[b])-2*logx c^3)]

Express the following as a single logarithm in simplest form.

(my math is rusty..., never good with fractions...)

edit:
i do have a question where values for x, y, z are given but not for logr and its asking to evaluate the expression...

4. Originally Posted by Jhevon
ok, not a problem. but did i get the question right? once i know that we can proceed
for 3^(2x-5) = 15^(2-x)
yes i believe its right (i like u explaining the steps..., i wish i would write my notes like that... this will be handy when exams start in 2 weeks)

5. Originally Posted by Jhevon
no, i mean the log question, did i type it out correctly? i didn't really explain all the steps for the first one, but if you got it, that's good

Yes that is typed out correct.

6. Originally Posted by jessyc
Yes that is typed out correct.
ok, good

Here are the rules you need to know for this problem.

1. $\log_a x^n = n \log_a x$

2. $\log_a xy = \log_a x + \log_a y$

3. $\log_a \left( \frac {x}{y} \right) = \log_a x - \log_a y$

$\frac {2}{3} \left[ 12 \log_x a - 6 \log_x \sqrt [4] {b} - 2 \log_x c^3 \right]$

$= \frac {2}{3} \left[ \log_x a^{12} - \log_x b^{ \frac {6}{4}} - \log_x c^6 \right]$ .......applied rule 1.

$= \frac {2}{3} \left[ \log_x \left( \frac {a^{12}}{b^{ \frac {3}{2}}} \right) - \log_x c^6 \right]$ ......applied rule 3 to the first two logs

$= \frac {2}{3} \left[ \log_x \left( \frac {a^{12}}{b^{ \frac {3}{2}} \cdot c^6} \right) \right]$ .......applied rule 3 including the last log

$= \log_x \left( \frac {a^{12}}{b^{ \frac {3}{2}} \cdot c^6} \right)^{ \frac {2}{3}}$ .......applied rule 1.

Now you can simplify this further if you wish (you should!)

You may be wondering, how come we never used rule 2, well, we could have but i decided not to.

7. Originally Posted by Jhevon
ok, good

Here are the rules you need to know for this problem.

1. $\log_a x^n = n \log_a x$

2. $\log_a xy = \log_a x + \log_a y$

3. $\log_a \left( \frac {x}{y} \right) = \log_a x - \log_a y$

$\frac {2}{3} \left[ 12 \log_x a - 6 \log_x \sqrt [4] {b} - 2 \log_x c^3 \right]$

$= \frac {2}{3} \left[ \log_x a^{12} - \log_x b^{ \frac {6}{4}} - \log_x c^6 \right]$ .......applied rule 1.

$= \frac {2}{3} \left[ \log_x \left( \frac {a^{12}}{b^{ \frac {3}{2}}} \right) - \log_x c^6 \right]$ ......applied rule 3 to the first two logs

$= \frac {2}{3} \left[ \log_x \left( \frac {a^{12}}{b^{ \frac {3}{2}} \cdot c^6} \right) \right]$ .......applied rule 3 including the last log

$= \log_x \left( \frac {a^{12}}{b^{ \frac {3}{2}} \cdot c^6} \right)^{ \frac {2}{3}}$ .......applied rule 1.

Now you can simplify this further if you wish (you should!)

You may be wondering, how come we never used rule 2, well, we could have but i decided not to.
wow, , if i read/learn and understand those rules, it should help me understand almost any question like that shouldn't it?

i read it fast, but it makes sense to me, and once again thanks for the comments on the side, they help

8. Originally Posted by jessyc
wow, , if i read/learn and understand those rules, it should help me understand almost any question like that shouldn't it?

i read it fast, but it makes sense to me, and once again thanks for the comments on the side, they help
yes, those are three of the four log rules that you MUST know. with those you should be able to do any log problem. When i tutor people at school i usually give them like 10 or so rules to memorize, it makes life a lot easier on them to just know those rules and apply them, but all the rules can be derived from the first four.

in case you are wondering, the fourth law which i mentioned (which really should be the first law, since it is the definition of a logarithm) is:

The logarithm of a number to a given base, is the power to which the base must be raised to give the number. That is,

If $\log_a b = c$,

then $a^c = b$

9. ## great help

Originally Posted by Jhevon
yes, those are three of the four log rules that you MUST know. with those you should be able to do any log problem. When i tutor people at school i usually give them like 10 or so rules to memorize, it makes life a lot easier on them to just know those rules and apply them, but all the rules can be derived from the first four.

in case you are wondering, the fourth law which i mentioned (which really should be the first law, since it is the definition of a logarithm) is:

The logarithm of a number to a given base, is the power to which the base must be raised to give the number. That is,

If $\log_a b = c$,

then $a^c = b$
yeah, that always helps me in class when the teacher actually goes through and shows where they derived from... although she didn't do this for logs, if she did i wasn't 'mentaly' there.

if your up to it, i have 2 more questions, but you really really don't have to!
(becuase where i am its really late, not sure about you)

one i should be able to get from the laws you shown me,

but the other is a question that is using the a(b)= x/c

10. this might sound kinda corny, but umm, i think you've made my night...
and some what inspired me to keep at this 'math' thing

edit:

heres the new topic...
http://www.mathhelpforum.com/math-help/urgent-homework-help/15615-b-x-b-x-getting-x-value.html#post54924

if you're up to the challange.