How would I expand the expression as a sum, difference, and/or constant multiple..... log7(xy^4/z^3) . I came up with 1/2[7logXY^4-7logz^3] not sure if I am on right track?

Question

How would I expand the expression as a sum, difference, and/or constant multiple.....    log7(xy^4/z^3)   . I came up with 1/2[7logXY^4-7logz^3]  not sure if I am on right track?

campbell_st · Answer

ok... here are a few laws you need to know about

1. $$\log(ab) = \log(a) + \log(b)$$

2. $$\log(x^a) = alog(x)$$

and lastly

$$\log(\frac{a}{b}) = \log(a) - \log(b)$$

ok so you question is 

$$\log7(\frac{xy^4}{z^3})$$

start by separating the 7 and the brackets... using the 1st log law shown above

anonymous · Answer

I see, so it would start with log7(xy^4)+log7(z^3) .. so far does that sound right?

campbell_st · Answer

not quite... I just want to check that

its not a base 7 log is it... ?

$$\log_{7}(\frac{xy^4}{x^3})$$

is it..?

anonymous · Answer

ummm, the way you have problem typed is correct regarding the position of the 7 next to the log

campbell_st · Answer

ok.... choice 1

$$\log_{7}(\frac{xy^4}{z^3})$$

or choice 2

$$\log7(\frac{xy^4}{z^3})$$

which choice... is the question..?

anonymous · Answer

choice 1 ... sorry about not placing the numbers correctly, I am still new to the site and trying to figure out how to format questions correctly :)

campbell_st · Answer

ok... that makes it easier... 

I use the division rule 1st

$$\log_{7} (xy^4) - \log_{7}(z^3)$$

separate the term in the 1st log using the multiplication rule... rule 1?

anonymous · Answer

gotcha, so I am thinking  $$\log_{7}(x)+\log_{7}(y ^{4})+\log_{7}(z ^{3})   $$

campbell_st · Answer

nearly... is

$$\log_{7}(x) + \log_{7}(y^4) - \log_{7}(z^3)$$

you just need to check your signs... 

last step... use the power rule... rule 2.. on the 2nd and 3rd logs

anonymous · Answer

ah I got the 3rd log mixed up but didn't need to do anything for the 2nd step. So the last step I am coming up with ... $$\log_{7}+4\log_{7}y-3\log_{7}z   $$

anonymous · Answer

not sure if I need the () around the letters y&Z still?

campbell_st · Answer

nearly just fix the 1st log... you left out the x..

campbell_st · Answer

I normally write the term in brackets... it avoids confusion

anonymous · Answer

oh duh, lol    Good point on the brackets   so the 1st log should be $$\log_{7}(x) $$  and of course the other logs am I correct?

campbell_st · Answer

yep

$$\log_{7}(x) + 4\log_{7}(y) - 3\log_{7}(z)$$

campbell_st · Answer

good work.

anonymous · Answer

makes so much more sense the way you explained it :)   I appreciate your time and explanation.  Thank you!