Question

Logarithms: how do I solve an equation when the bases are different, example:
log4(x) - log8(x) = 1

Answer 1

log8(x) = log4(x) / log4(8)

Answer 2

log4(8) = log4(2*4) = log4(2) + log4(4) = 1/2 + 1 = 3/2

Answer 3

change the base of the log8(x) like i did above

Answer 4

yes...replace log8(x) with that...and then use normal rules for logs to solve.

Answer 5

to make things easier, multiply through by a 3

Answer 6

why don't you guide him to make 1 = log 4 of 4 , too? finish the stuff to make it clear., please

Answer 7

i don't think making 1=log4(4) helps, really.

and i'd like to see how far he can take it without a different base.

but one more step might spark things.

once you replace and multiply through by 3, you have

3log4(x) - 2log4(x) = 3

you will need to use two rules:
1. a*log(x) = log(x^a)
2. log(a) - log(b) = log(a/b)

use those two rules and then simplify...you'll get a log on one side and a number on the other

Answer 8

you arrive at the same place in the end. I'm not sure fractions are easier, IMO...

Answer 9

definitely, @loser66...two ways to get to the same end. but i've seen many students who abhor fractions enough...especially fractional exponents :)

I try to avoid them if possible since these can be done multiple ways. and the log rules used are the same in both cases.

Answer 10

in my class, i would address fractions as fractions...if they don't get logs and have issues with fractions, you just compound problems if you put them together. i would attack the issues separately so they are individually clear.

fractions alone are issues for students, but then you add in the fractional exponents are roots...and then you have these things called logs? that is enough to shut down some students

Answer 11

@fsateles it's rarely to meet a math prof. try to get something from him, ask whatever you don't know, most important is : if consider the hard problem as your "enemy", "operate" it, and kill it, don't let it threaten you. hehehe.. sorry for my aggressiveness.

Answer 12

I did not understood how you got it: 3log4(x) - 2log4(x) = 3

Answer 13

haha...no apology needed :) one of the best things about math is there are usually enough weays to get to the end that you can look at it whichever way your brain takes you best.

Answer 14

do you understand that log8(x) = (2/3)*log4(x)?

Answer 15

yes.

Answer 16

but why it turns to 3log4(x) - 2log4(x) = 3?

Answer 17

i multiplied through the equation by 3

Answer 18

i'm really confused o.o. Look: I got this point
\[\log_{4}x - (2 \log_{4}x \div 3) = 1\]

Answer 19

ok, so if you multiply each term by 3, to clear the fraction, what do you get?

Answer 20

3 * log4(x) - 3 * (2 * log4(x) / 3) = 3 o.o

Answer 21

so the middle term...the 3 on top cancels with the 3 on the bottom, right?

Answer 22

leaving 2 * log4(x)

Answer 23

ahhh!! Of course. I'm a dumb. -_-"

Answer 24

Sorry for that and thank you very much for patience.

Answer 25

lol...no problem. not dumb...sometimes the simple things are obvious