# logarithms questions

• Sep 29th 2008, 07:42 AM
whatsgoingon
logarithms questions
could somebody help me with these questions...being able to see the work would be awesome and incredibly helpful! thank you!

1) write as a single logarithm and solve :
2log3(12)-2log3(4)

2)write as a single logarithm:
log3(x+2)-logx(2x-1)+log3(5)

3)solve for x: logx(x-1)=log8(x-1)
• Sep 29th 2008, 09:57 AM
wisterville
Hello,

You should know some formulae for logarithms. Then, 1) and 2) are easy.

For 3), you see that x-1>0 (in order for the log to be defined.)
Then, the equation becomes $\displaystyle \frac{\log_8(x-1)}{\log_8x}=\log_8(x-1)$.

Bye.
• Sep 29th 2008, 11:02 AM
GoneCrazy
HEY,

3) Since the question is asking u to solve to find x
then x has a value
just looking at the eqation you can conclude that
(x=8), since its similar and there is an = sign
soooo the bases have to be the same too.

For 1 and 2, dude its veryyyyy simple(Surprised)
you just need to know 4 basic rules.

1) nLog x = Log x^n
2) Log(base-C)A + Log(base-C)B = Log(baseC)A*B
3) Log(base-C)A - Log(base-C)B = Log(baseC)A/B
4) Log (base-x)x^n = n Log (base-x)x =n

* = times

Back to ur questions

1) 2log3(12) - 2log3(4)
if i am correct 3 is the base????

Log (base3) 12^2 - Log (base3) 4^2
Log (base3) 144 - Log (base3) 16
Log (base3) 144/16
Log (base3) 9 (thats the single logarithm)

We can further simplify it to solve it
Log (base3) 3^2
2 Log (base3) 3
= 2

2) Log 3(x+2) - Log x(2x-1) + log 3(5)
= Log (3x+6) - log (2x^2 -x) + log 15

= log (3x+6)/(2x^2 -x) * 15
= log (45x+90)/(2x^2 -x)

I hope this helpes
cheers
• Sep 29th 2008, 12:12 PM
Simplicity
Quote:

Originally Posted by whatsgoingon
could somebody help me with these questions...being able to see the work would be awesome and incredibly helpful! thank you!

1) write as a single logarithm and solve :
2log3(12)-2log3(4)

2)write as a single logarithm:
log3(x+2)-logx(2x-1)+log3(5)

3)solve for x: logx(x-1)=log8(x-1)

Here are the general rules. Using these should enable you to solve your question. You may need to use more than one rules for some of your questions.

$\displaystyle a\log_x(y)=\log_x(y^a)$

$\displaystyle \log_x(y) + \log_x(z) = \log_x(yz)$

$\displaystyle \log_x(y) - \log_x(z) = \log_x\left(\frac{y}{z}\right)$

$\displaystyle \log_x(y) = \frac{\log_z(y)}{\log_z(x)}$

$\displaystyle a^y = x \rightarrow y=\log_a(x)$