# bunch of Logarithm qusetions help^^

• Jan 6th 2009, 08:01 PM
hovermet
bunch of Logarithm qusetions help^^
If log3 4=X express each of the following in terms of x

a) log3 16-2log3 32 b)log3 2

Express y as a function of x. state the domain
a)logY = log(1-3x)
b)log2 (y-3) = 5 + log2 (x-4)

solve and check
a)2logM + 3logM =10

Hi i am totally stuck with those log questions
can anyone show me how to do it in full steps?

i tried the last one 2logM + 3logM =10 i got stuck after moving the 2 to the power of M.

for the other ones i dont have any idea on how to start on it..ty
• Jan 6th 2009, 08:39 PM
Re :
(1) Given $\displaystyle log_3 4=x$

(a) $\displaystyle log_3 16-2log_3 32$
$\displaystyle = log_3 4^2-2log_3 2^5$
$\displaystyle = log_3 4^2-log_3 2^{10}$
$\displaystyle =log_3 4^2-log_3 (4^\frac{1}{2})^{10}$
Can you continue from here ?

(b) $\displaystyle log_3 2=\frac{1}{2}log_3 4$ = 1/2x

(2) $\displaystyle lg y= lg(1-3x)$
Cancel logs from both sides , then $\displaystyle y=1-3x$

$\displaystyle log_2 (y-3)=5+log_2(x-4)$
$\displaystyle log_2 (y-3)=log_2 2^5+log_2(x-4)$
$\displaystyle log_2 (y-3)=log_2(2^5(x-4))$
Take away logs from both sides .....

(3) $\displaystyle 2lg M +3lg M =10$
$\displaystyle lgM^2+lgM^3=10$
$\displaystyle lgM^5=10$
Try continue from here ...
• Jan 6th 2009, 10:13 PM
hovermet
1)

a)u cancel the logs than turn 4^(1/2)^10 into 4^5?
then 4^(2-5) like this? Then i got stuck again >.< sry

and for B how did u get 1/2x right away?

2)i dont understand what "y as a function of x" means. What are you actually looking for?

3)
$\displaystyle$
$\displaystyle lgM^5=10$

5=M^10
then i got stuck again>.< do you square root the 5 10 times?

>.<ty(Nerd)
• Jan 6th 2009, 11:17 PM
earboth
Quote:

Originally Posted by hovermet
1)

a)u cancel the logs than turn 4^(1/2)^10 into 4^5?
then 4^(2-5) like this? Then i got stuck again >.< sry

and for B how did u get 1/2x right away?

2)i dont understand what "y as a function of x" means. What are you actually looking for?

3)
$\displaystyle$
$\displaystyle lgM^5=10$

5=M^10
then i got stuck again>.< do you square root the 5 10 times?

>.<ty(Nerd)

1. Keep in mind that $\displaystyle \boxed{x = \log_3(4)}$

2. From

$\displaystyle =log_3 4^2-log_3 (4^\frac{1}{2})^{10} = \log_3(4^2)-\log_3{4^5} = 2\log_3(4)-5\log_3(4) = 2x-5x = -3x$

3. I guess 1)(b) is obvious right now. (See my initial remark)

4. You are asked to derive an equation where y is expressed by x. In other words: Solve the given equation for y and look what happens to the x.

a)$\displaystyle \log Y = \log(1-3x)$ The base of these logs isn't specified so you can use any base. (Mathaddict has used the base 10). Use the logs as exponents to a base and you'll get:

$\displaystyle y = 1-3x$ This equation describes a function y of x.

b) $\displaystyle \log_2 (y-3) = 5 + \log_2 (x-4)$ The base is 2. Use this base and the given logarithms (=exponents) to get rid of the logarithms:

$\displaystyle 2^{\log_2 (y-3)}=2^{5 + \log_2 (x-4)} ~\implies~y-3 = 2^5 \cdot (x-4)~$ $\displaystyle \implies~y-3=32x-128~\implies~\boxed{y=32x-125}$

solve and check
a) $\displaystyle 2 \log M + 3 \log M =10~\implies~5\log M =10~\implies~\log M = 2$

If you use $\displaystyle \log (n)= \log_{10}(n)$ then the equation becomes: $\displaystyle \boxed{M = 100}$
• Jan 7th 2009, 08:28 PM
hovermet
ty>.<

but i still don't understand the first question

how did u get 2\log_3(4)-5\log_3(4) = 2x-5x = -3x? where did the X came from and can't u just add the exponents? since the have the same base. also on my text the answer is suppose to be 1 + 3X why?

and i got another question

log3 2 = x (same as question one

a)log3 8

i got the the point where log8/log3 =x after i done that i got 1.8xxxxx. but the answer is 3X..
============

log3 4=x

a)log3 16- 2log3 32
i got to the point where 3log3 4 - 8log3 4 and came out with an answer -5X, but in the text it saids the answer is -3X....help>.<
================
slove for X
log8 x-1 = log8 (x-1)

i canceled log8 out from both sides and got x-1 = x-1...... ?= -2 stuck...why can't u just cancel both sides and find the answer>.< log is so complicated>.< help
====================
solve and check

log4 (x+2) + log4 ( x-1) =1
got stuck when i got to x^2+ x-3=0 can't factor..

• Jan 7th 2009, 10:09 PM
earboth
Quote:

Originally Posted by hovermet
If log3 4=X express each of the following in terms of x

...

Quote:

Originally Posted by hovermet
ty>.<

but i still don't understand the first question

how did u get 2\log_3(4)-5\log_3(4) = 2x-5x = -3x? where did the X came from and can't u just add the exponents? since the have the same base. also on my text the answer is suppose to be 1 + 3X why?

...

You have given this term in your first post :D
• Jan 7th 2009, 10:14 PM
earboth
Quote:

Originally Posted by hovermet
...
log3 2 = x (same as question one

a)log3 8

i got the the point where log8/log3 =x after i done that i got 1.8xxxxx. but the answer is 3X..
============

...

If $\displaystyle \log_3(2) = x$ then

$\displaystyle \log_3(8)=\log_3(2^3)=3\log_3(2) = 3x$

(and don't ask me where the x comes from ... !)
• Jan 7th 2009, 10:26 PM
earboth
Quote:

Originally Posted by hovermet
...
log3 4=x

a)log3 16- 2log3 32
i got to the point where 3log3 4 - 8log3 4 and came out with an answer -5X, but in the text it saids the answer is -3X....help>.< I answered this question already. See post #4 of this thread.
================
slove for X
log8 x-1 = log8 (x-1)
i canceled log8 out from both sides and got x-1 = x-1...... ?= -2 stuck...why can't u just cancel both sides and find the answer>.< log is so complicated>.< help
====================

$\displaystyle \log_8(x) - 1 = \log_8(x-1)~\implies~x\cdot \frac18=x-1~\implies~-\dfrac78 x=- 1~\implies~x= \dfrac87$

Quote:

Originally Posted by hovermet
solve and check

log4 (x+2) + log4 ( x-1) =1
got stuck when i got to x^2+ x-3=0 can't factor..

$\displaystyle \log_4 (x+2) + \log_4 ( x-1) =1~\implies~(x+2)(x-1)=4~$ $\displaystyle \implies~x^2+x-6=0~\implies~x=-3~\vee~x=2$

Since the domain of this equation is $\displaystyle d=(1, \infty)$ x = -3 isn't a solution of the equation.