Let log_bA=3 log_bC=2 log_bD=5

then what is the value of
log_b D^2/ C^3 A

Question

Let log_bA=3  log_bC=2  log_bD=5

then what is the value of
log_b D^2/ C^3 A

anonymous · Answer

$$\log_b \frac{ D^2 }{ C^3A }$$

anonymous · Answer

thats what i am looking for the value for ^^^^^

mathmate · Answer

Use the laws of logarthms:
$$log(ab)=log(a)+log(b)$$
$$log(a^b)=b\ log(a)$$
to simplify the expression and substitute the given values.

anonymous · Answer

mmmhhh ok then

geerky42 · Answer

He forgot one more: $\log\left(\dfrac{a}{b}\right) = \log(a) - \log(b)$

You would need to apply that first. Can you do it?

anonymous · Answer

mmmhhh i am attempting right now .... lol would you mind checking if i got it when im done ??? @geerky42

geerky42 · Answer

Sure

mathmate · Answer

@geerky42  yep, good catch!

anonymous · Answer

lol soooo i did all of the properties and something went wrong my answer was not one of the answers provided .... heres what i did 

$$\log_b \frac{ D^2 }{ C^3 A }$$

$$\log_bD^2-\log_bC^3A$$

$$2\log_bD-\log_bC^3+\log_bA$$

$$2\log_bD-3\log_bC+\log_bA$$

from here i simply added the numbers ... 
2(5) - 3(2) + 3 
10-6+3 = 7

anonymous · Answer

@geerky42  @mathmate  okkk technical difficulties .... -_- lol

mathmate · Answer

I think the sign for log(A) should be negative...

mathmate · Answer

... unless you add parentheses around (3log(C)+log(A))

geerky42 · Answer

+logbA is error here.$$-\log_bC^3A = -(\log_bC^3A) = -(\log_bC^3+\log_bA) = -\log_bC^3 -\log_bA$$

anonymous · Answer

ohhhh didnt apply the negativeeeee ohhh ok

anonymous · Answer

ohh then its one !

geerky42 · Answer

Right!

anonymous · Answer

yay thats actually an answer given thankkkk you n_n !!@geerky42 eheheh i did it n_n

anonymous · Answer

thank you n_n @geerky42 @mathmate

mathmate · Answer

You're welcome!  :)