# Thread: logarithms questions

1. ## 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)

2. 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.

3. 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
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

4. 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.

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

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

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

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

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