Help with logarithms?

This is a common problem with BIDMAS (or PEMDAS where I live). You seem to have performed addition before subtraction "because A is before S".

The A and S in BIDMAS go together: "Add and Subtract, left-to-right". So 5 - 2 + 3 groups as (5 - 2) + 3 = 6, not as 5 - (2 + 3) = 0.

You must know that from earlier courses, so the newness of logarithms must be getting in the way.

3 log 2 - log 6 + log 12 = log 8 - log 6 + log 12 = log (8/6) + log 12 = log (8/6 * 12) = log 16

All that works for any logarithm, base 4 or other. Given base 4, you can finally simplify:

log_4 16 = 2

3 log₄(2) - log₄(6) + log₄(12)
= log₄(2³) - log₄(6) + log₄(12)
= log₄(8) - log₄(6) + log₄(12)
= log₄(8) + log₄(12) - log₄(6)
= log₄(8 × 12 / 6)
= log₄(16)
= log₄(4²)
= 2 log₄(4)
= 2


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Why did you get a wrong answer ?

3 log₄(2) - log₄(6) + log₄(12)
≠3 log₄(2) - [log₄(6) + log₄(12)] …… It should be 3 log₄(2) - [log₄(6) - log₄(12)] instead
= log₄(2³) - log₄(6 ×12)
= log₄(8) - log₄(72)

Hence, 3 log₄(2) - log₄(6) + log₄(12) ≠log₄(8) - log₄(72)

General rules:
n·log(x) = log(xⁿ)
logᵪ(xⁿ) = n
log(ab) = log(a) + log(b)
log(a/b) = log(a) - log(b)

3log₄(2) - log₄(6) + log₄(12) = 2
 

3log₄2 - log₄6 + log₄12

Recall from the rules of logarithm:
n·log(x) = log(xⁿ) {Power Rule}

Then, we calculate thus:
3log₄2 - log₄6 + log₄12
= log₄2³ - log₄6 + log₄12
= log₄8 - log₄6 + log₄12
= log₄8 + log₄12 - log₄6 {from BODMAS}

Recall from the rules of logarithm:
log (a) + log (b) = log (ab) {Product Rule}

Then,
log₄8 + log₄12 - log₄6
= log₄(8 × 12) - log₄6
= log₄96 - log₄6

Recall from the rules of logarithm:
log (a) - log (b) = log (a/b) {Quotient Rule}

Then,
log₄96 - log₄6
= log₄(96/6)
= log₄16
= log₄4²
= 2log₄4

Recall from the rules of logarithm:
logᵪx = 1

Then,
2log₄4
= 2(1)
= 2 ...Ans.

Presuming you mean:

3 log₄(2) - log₄(6) + log₄(12)

The first thing that I'll do is move the 3 inside the first log as an exponent:

log₄(2³) - log₄(6) + log₄(12)
log₄(8) - log₄(6) + log₄(12)

Now we use the sum/difference of two logs of the same base rule to turn this into one log of a product/quotient:

log₄(8 * 12 / 6)

Simplify:

log₄(8 * 2)
log₄(16)

16 is a power of 4, so:

log₄(4²)

And now that you have a number of base 4 with a log of base 4, they cancel each other out leaving only the exponent:

2

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When you got to log₄(8) - log₄(72), you violated the order of operations. Addition and subtraction are done at the same level from left to right. Since you did the addition first, you ended up subtracting the last term instead of adding it. That's why things failed.

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If this helped, please give best answer.

3log(base 4)2 - log(base 4)6 + log(base 4)12 =
log(base 4)8 - log(base 4)6 + log(base 4)12 =
log(base 4)(8/6 * 12) =
log(base 4)16 =
2

3log4(2)-log4(6)+log4(12)

3 * log[4](2) - log[4](6) + log[4](12)

Let's just say that this is a - b + c. How on earth did you switch the + sign into a - sign in front of c?

log[4](2^3 12 / 6) =>
log[4](8 * 2) =>
log[4](16) =>
log[4](4^2) =>
2 * log[4](4) =>
2 * 1 =>
2

Let log be log to base 4

log 8 - log 6 + log 12 = log 96 - log 6

log 96 - log 6 = log 16

log 16 = 2

Maybe if you had
log4(8) - (log4(6) + log4(12))
Then log4(8) - log4(72) is correct because the log4(6) + log4(12) is grouped together.

log4(8) - log4(6) + log4(12)
log4(8) + log4(12) - log4(6)
Now if you want first combine the addition into a product
log4(8*12) - log4(6)
And now turn the subtraction into division
log4(8*12 / 6)
= log4(16)
= 2

Simply, all arguments of + logs will end up being multiplied in the final log, and you divide the product with all the arguments of - logs.

(2^3/6)*(12)=16

So far so good!!!!
Then
log(4) [8/72] =>
log4[9] =.
2log4[3]