Question

Log_{4}8 + Log_{4}64/3 - Log_{4}8
apparently the answer is 2...but i'm getting 3..

Answer 1

Sorry, Evaluate:^

Answer 2

= log_{4}( (8 * 64) / (3 * 8))
=log_{4}(64 / 3)
=log_{10}(64/3) / log_{10}(4)
= 2.2

Answer 3

unless the second term is [Log_{4} (64) ]/3 then its a different answer, i probably read it wrong

Answer 4

the left and right log's cancel out so u are left with log_[4](64/3) now to get it into log_10 form divide 64/3 and then raise it to the 1/4 power which gives 2.1

Answer 5

what is the answer exactly?

Answer 6

Back of the mathpower 12 book just says 2 even.
I came to 3 this way.
I multiplied all by 3 to kill the fraction and ended up with
Log_{4}24 + Log_{4}64 - Log_{4}24
Log_{4}(24)(64)/24
Log_{4}64
Log_{4}4^3
3Log_{4}4 = 3

was there something in there I am not supposed to be doing?

Answer 7

pretty sure you can't just multiply it by three to get eliminate the fraction because then there would be nothing stopping you from multiplying each term by, say, 3000 and getting a completely different answer since the first and last terms would still cancel

Answer 8

secondly, multiplying each term by three like 3 * log_{4}(64/3) doesn't equal log_{4}(64), it is log_{4} ((64/3)^3)
if you want to remove the fraction you have to add log_{4}(3) to log_{4}(64/3) so the 3s cancel.
my guess is the book just rounded the answer down

Answer 9

Is this your question?\[\log_{4}8 + \log_{4}\frac{64}{3} - \log_{4}8\]

Answer 10

Another hypothesis. Is this your question?\[\log_{4}8 + \frac{\log_{4}64}{3}-\log_{4}2\]Notice where I put the 3, and I changed the last 8 to a 2.

Answer 11

Back of the mathpower 12 book just says 2 even.
I came to 3 this way.
I multiplied all by 3 to kill the fraction and ended up with
Log_{4}24 + Log_{4}64 - Log_{4}24
Log_{4}(24)(64)/24
Log_{4}64
Log_{4}4^3
3Log_{4}4 = 3

was there something in there I am not supposed to be doing?

Answer 12

the first one is the question exactly. Sorry for the late response, the website was actin a fool. lol

Answer 13

Then mike07 had it right in the 2nd response.

Answer 14

He didn't even have to do the first step because the first and last terms sum to 0, so they disappear.

Answer 15

\[\log_{4}8+\log_{4}\frac{64}{3}-\log_{4}8=\log_{4}\frac{64}{3}=\frac{\log(64/3)}{\log4}\]Nothing left to do but use a calculator.