A question with Logarithms?

Question

I am teaching myself maths and am attempting some exam questions in the book, I have managed all the others but this one has stumped me. So here goes

.

It says, Explaining each step claearly express

By the way, all logs are to the base 2

so Log(8 * 3^1/2) - 1/3log9/16

Express in the form p + qlog3 where p and q are rational numbers to be found.

I know the laws of logs, but if you could show each step that would be great.
The answer in the back is 13/3 - 1/6log3

Happy new year everyone too

Answer 1

First step is to expand the given log by using log properties...

let's take the first part log(8*3^1/2)...
log(a*b) = log(a) + log(b)... log identity
log[8*3^(1/2)] = log8 + log3^(1/2)

log a^b = b*log a... log identity
log 3^(1/2) = (1/2)*log 3

log 8 can be re written as log 2^3... apply the same identity we just used... now we have... 3 * log 2...
log a to the base a = 1

and you said all logs are to the base 2, so...
log 2 to the base 2 = 1
log 8 = log 2^3 = 3*log 2 = 3...

so our final first part is... 3 + (1/2)log 3

let's do the second part now... (-1/3) log 9/16
log a/b = log a - log b
so (-1/3)log 9/16 = (-1/3) [log 9 - log 16]
once we open the brackets, we have...
-1/3 log 9 + 1/3 log 16
-1/3 log 9 can be re written as... -1/3 log 3^2
again use log a^b = b*log a identity and you have... -2/3 log 3

1/3 log 16 can be rewritten as... 1/3 log 2^4... 4/3 log 2... but we said earlier that log 2 to the base 2 is 1, so 4/3 log 2 = 4/3

final second part will be... 4/3 - 2/3 log 3

now put everything back to its original place...

3 + (1/2) log 3 + 4/3 - (2/3) log 3
3 + 4/3 = 13/3
1/2 log 3 - 2/3 log 3 = -1/6 log 3

ur final answer is 13/3 - 1/6 log 3...

these kind of problems need manipulation... if you know the log identities, you can easily manipulate them... ;)

Answer 2

Very Easy
log (8 x 3^1/2) / (9/16)^1/3

= log {[(2^3) x (3^1/2)] / [(9^1/3) x 16^1/3]}
= log {[(2^3) x (3^1/2) x (16^1/3)] / (9^1/3)}
= log {[(2^3) x (3^1/2) x (2^4/3)] / (3^2/3)}
= log [(2^3) x (2^4/3)] + log [(3^1/2) / (3^2/3)]
= log [2^(3+4/3)] + log [3^(1/2 - 2/3)]
= log 2^(13/3) + log 3^(-1/6)
= 13/3 log 2 + -(1/6) log 3

Log 2 to the base 2 = 1
So,
= 13/3 -1/6log 3

The thing u have to keep in mind is to reduce the number as far as u can by making it power.
like 16 = 2^4 9=3^2

Answer 3

(log 8*3^1/2) - (1/3)(log 9/16)

Using the rule that log a + log b = log ab,

(log 8)+(log 3^1/2) - ((1/3)(log 9) - (1/3)(log 16))

Power rule, (remember that 2^3=8, and 2^4=16)

3+(log 3)/2 - (2(log 3)/3 - 4/3)

3 + (log 3)/2 - 2(log 3)/3 + 4/3

13/3 + (log 3)/2 - 2(log 3)/3

13/3 + (log 3)(1/2-2/3)

13/3 + (log 3)*(-1/6)

13/3 - (1/6)(log 3)

Answer 4

log(8*3^1/2) = log(8) + log(3^1/2) = 3 + (1/2)log(3)

(1/3)log(9/16) = (1/3)log(9) - (1/3)log(16) = (1/3)log(3^2) - 4/3 = (2/3)log(3) - 4/3

Subtracting the two, we get:
log(8*3^1/2) - (1/3)log(9/16) = 3 + (1/2)log(3) - (2/3)log(3) + 4/3 = 13/3 - (1/6)log(3)

Answer 5

=log 8+log 3^1/2-1/3(log 9-log 16) all logs are to the base 2
=log2^3+log 3^1/2-1/3(log 3^2-log 2^4)
=3log 2+1/2log3- 1/3(2log3-4log2)
=3log2+1/2log3-2/3log3+4/3log2 we know that log2=1
=3*1+1/2log3-2/3log3+4/3*1
=3+1/2log3-2/3log3+4/3
=(3+4/3)+(1/2-2/3)log3
=13/3-1/6 log 3