# Please someone help me on how to tackle this question. Am stuck for days.?

(3log√2 + log24log3) / (log512-½log64)

### 12 Answers

- ?Lv 72 months ago
log(a^b) = b log(a)

Write each expression in term of log(3) and log(2).

For example, log(√2) = 0.5 log(2)

- PinkgreenLv 72 months ago
The original expression=

{1.5log(2)+log(3)[log(3)+3log(2)]}/[9log(2)-3log(2)]=

{2[log(3)]^2+6log(2)log(3)+3log(2)}/[12log(2)]=

{[log(3)]^2}/[6log(2)]+log(3)/2+1/4~

0.614597176.

{Note that: log[x] is usually considered log(10)[x]!}

You may use your calculator directly from beginning

since not much simplification can be made.

To the one who objected my result, What wrong it was?

You should check carefully over the problem before posting it;

you must not make any silly conclusion without reason!

- llafferLv 72 months ago
Not sure what the actual question is. If this is just to simplify it, let's see what we can do with this:

[ 3 log(√2) + log(24) * log(3) ] / [ log(512) - (1/2)log(64) ]

The first thing that I see is 3 of the 5 numbers are all powers of 2. 24 is made up of the product of powers of 2 and 3. So let's start out by getting the prime factorization of everything:

[ 3 log(2^0.5) + log(2³ * 3) * log(3) ] / [ log(2⁹) - (1/2)log(2⁶) ]

Next, that log of a product in the second term in the numerator can be simplified to be the sum of two logs:

[ 3 log(2^0.5) + [ log(2³) + log(3) ] * log(3) ] / [ log(2⁹) - (1/2)log(2⁶) ]

Now every exponent inside a log can be pulled out of it and simplified with any pre-existing coefficient to that log:

[ 3 * 0.5 log(2) + [ 3 log(2) + log(3) ] * log(3) ] / [ 9 log(2) - (1/2) * 6 log(2) ]

[ 1.5 log(2) + [ 3 log(2) + log(3) ] * log(3) ] / [ 9 log(2) - 3 log(2) ]

The denmoniator now has two like terms that can be simplified:

[ 1.5 log(2) + [ 3 log(2) + log(3) ] * log(3) ] / [ 6 log(2) ]

Let's expand the log multiplied by the sum of two logs:

[ 1.5 log(2) + 3 log(2) log(3) + log²(3) ] / [ 6 log(2) ]

We have the sum of three terms over a denominator. We can split this up into the sum of three fractions with a common denominator:

1.5 log(2) / [ 6 log(2) ] + 3 log(2) log(3) / [ 6 log(2) ] + log²(3) / [ 6 log(2) ]

We can now cancel out common log(2)'s and simplify the rational coefficients:

1.5 / 6 + 3 log(3) / 6 + log²(3) / [ 6 log (2) ]

1/4 + log(3) / 2 + log²(3) / [ 6 log (2) ]

And as one more step to try to simplify this, we can turn the quotient of two logs with the same base into a single log with a modified base. We'll need to split up the squared log into separate logs multiplied together so we can apply it to only one of them:

1/4 + (1/2) log(3) + (1/6) log₂(3) * log(3)

Without going into decimal approximations, I think this is as simplified as we can get that original expression.

For people who are just spitting out 1.089516 or something equivalent, these are using base "e" logs, not base 10 logs. The actual decimal approximation is equal to 0.614597 (rounded to 6DP). You likely entered the expression into wolframalpha and you took the answer blindly without realizing that if you give it log(10) it won't give you 1 but instead will give you 2.302585 (rounded to 6DP).

-------------

Looking through other answers who think you may be missing a subtraction and instead should be:

[ 3 log(√2) + log(24) - log(3) ] / [ log(512) - (1/2)log(64) ]

We will start the same by getting the prime factorization of all terms:

[ 3 log(2^0.5) + log(2³ * 3) - log(3) ] / [ log(2⁹) - (1/2)log(2⁶) ]

Turn the log of the product into the sum of two logs:

[ 3 log(2^0.5) + log(2³) + log(3) - log(3) ] / [ log(2⁹) - (1/2)log(2⁶) ]

Pull out any exponents out of the log and simplify:

[ 3 * 0.5 log(2) + 3 log(2) + log(3) - log(3) ] / [ 9 log(2) - (1/2) * 6 log(2) ]

[ 1.5 log(2) + 3 log(2) + log(3) - log(3) ] / [ 9 log(2) - 3 log(2) ]

Now we can combine like terms: log(3) in the numerator and log(2) in the denominator:

[ 1.5 log(2) + 3 log(2) ] / [ 6 log(2) ]

Let's move the updated coefficients back into the logs as exponents:

[ log(2^1.5) + log(2³) ] / log(2⁶)

The sum of two logs is the same as the log of the product:

log(2^1.5 * 2³) / log(2⁶)

Multiplying two numbers with the same base is the same as adding exponents:

log(2^4.5) / log(2⁶)

Pull the exponents out of the logs one more time:

4.5 log(2) / 6 log(2)

And finally, we can cancel out the common log(2)'s and simplify what's left:

4.5/6

3/4

In this case, since all logs cancel out it doesn't matter what base the logs are, you get the same result either way.

- PhilomelLv 72 months ago
(3 log(sqrt(2)) + log(24) log(3))/(log(512) - log(64)/2) = (-6 i π floor(arg(2 - x)/(2 π)) + 8 π^2 floor(arg(3 - x)/(2 π)) floor(arg(24 - x)/(2 π)) - 3 log(x) - 4 i π floor(arg(3 - x)/(2 π)) log(x) - 4 i π floor(arg(24 - x)/(2 π)) log(x) - 2 log^2(x) + 3 sum_(k=1)^∞ ((-1)^k (2 - x)^k x^(-k))/k + 4 i π floor(arg(24 - x)/(2 π)) sum_(k=1)^∞ ((-1)^k (3 - x)^k x^(-k))/k + 2 log(x) sum_(k=1)^∞ ((-1)^k (3 - x)^k x^(-k))/k + 4 i π floor(arg(3 - x)/(2 π)) sum_(k=1)^∞ ((-1)^k (24 - x)^k x^(-k))/k + 2 log(x) sum_(k=1)^∞ ((-1)^k (24 - x)^k x^(-k))/k - 2 sum_(k_1 =1)^∞ sum_(k_2 =1)^∞ ((-1)^(k_1 + k_2) (3 - x)^(k_1) (24 - x)^(k_2) x^(-k_1 - k_2))/(k_1 k_2))/(2 i π floor(arg(64 - x)/(2 π)) - 4 i π floor(arg(512 - x)/(2 π)) - log(x) - sum_(k=1)^∞ ((-1)^k (64 - x)^k x^(-k))/k + 2 sum_(k=1)^∞ ((-1)^k (512 - x)^k x^(-k))/k) for x<0

(3 log(sqrt(2)) + log(24) log(3))/(log(512) - log(64)/2) = (3 floor(arg(2 - z_0)/(2 π)) log(1/z_0) + 2 floor(arg(3 - z_0)/(2 π)) floor(arg(24 - z_0)/(2 π)) log^2(1/z_0) + 3 log(z_0) + 3 floor(arg(2 - z_0)/(2 π)) log(z_0) + 2 floor(arg(3 - z_0)/(2 π)) log(1/z_0) log(z_0) + 2 floor(arg(24 - z_0)/(2 π)) log(1/z_0) log(z_0) + 4 floor(arg(3 - z_0)/(2 π)) floor(arg(24 - z_0)/(2 π)) log(1/z_0) log(z_0) + 2 log^2(z_0) + 2 floor(arg(3 - z_0)/(2 π)) log^2(z_0) + 2 floor(arg(24 - z_0)/(2 π)) log^2(z_0) + 2 floor(arg(3 - z_0)/(2 π)) floor(arg(24 - z_0)/(2 π)) log^2(z_0) - 3 sum_(k=1)^∞ ((-1)^k (2 - z_0)^k z_0^(-k))/k - 2 floor(arg(24 - z_0)/(2 π)) log(1/z_0) sum_(k=1)^∞ ((-1)^k (3 - z_0)^k z_0^(-k))/k - 2 log(z_0) sum_(k=1)^∞ ((-1)^k (3 - z_0)^k z_0^(-k))/k - 2 floor(arg(24 - z_0)/(2 π)) log(z_0) sum_(k=1)^∞ ((-1)^k (3 - z_0)^k z_0^(-k))/k - 2 floor(arg(3 - z_0)/(2 π)) log(1/z_0) sum_(k=1)^∞ ((-1)^k (24 - z_0)^k z_0^(-k))/k - 2 log(z_0) sum_(k=1)^∞ ((-1)^k (24 - z_0)^k z_0^(-k))/k - 2 floor(arg(3 - z_0)/(2 π)) log(z_0) sum_(k=1)^∞ ((-1)^k (24 - z_0)^k z_0^(-k))/k + 2 sum_(k_1 =1)^∞ sum_(k_2 =1)^∞ ((-1)^(k_1 + k_2) (3 - z_0)^(k_1) (24 - z_0)^(k_2) z_0^(-k_1 - k_2))/(k_1 k_2))/(-floor(arg(64 - z_0)/(2 π)) log(1/z_0) + 2 floor(arg(512 - z_0)/(2 π)) log(1/z_0) + log(z_0) - floor(arg(64 - z_0)/(2 π)) log(z_0) + 2 floor(arg(512 - z_0)/(2 π)) log(z_0) + sum_(k=1)^∞ ((-1)^k (64 - z_0)^k z_0^(-k))/k - 2 sum_(k=1)^∞ ((-1)^k (512 - z_0)^k z_0^(-k))/k)

=1.08952

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- Wayne DeguManLv 72 months ago
Judging by the space between log24 and log3, I think you are missing something vital. After playing around with this I think you mean log24 - log3

Now, using log(A/B) = logA - logB we have:

log24 - log3 = log8 => log2³ = 3log2

Now, 3log√2 = log(√2)³ => log(2³/²) = (3/2)log2

so, the numerator gives (3/2)log2 + 3log2 = (9/2)log2

Considering the denominator we have:

log512 = log(2⁹) => 9log2

and (1/2)log64 => log(64)¹/² = log8 => log2³ = 3log2

Hence, log512 - (1/2)log64 = 9log2 - 3log2 = 6log2

Dividing numerator by denominator we get:

(9/2)log2 ÷ 6log2 => (9/2) ÷ 6 = 3/4

Nice question.

:)>

- KrishnamurthyLv 72 months ago
(3 log(sqrt(2)) + log(24) log(3))/(log(512) - 1/2 log(64))

= (log(8) + log(9) log(24))/log(4096)

= 1.08951602439578815398796980513290686...

- MorningfoxLv 72 months ago
The way you wrote it, the result is 1.089516... The exact form is complicated. And if the logs are base 10, then the result is 0.614571...

But if there is a minus sign between log 24 and log 3, then the result is 3/4. It doesn't matter what the base is, because that part factors out.

Usually school problems have simple answers.

- PuzzlingLv 72 months ago
Usually after working on a problem for days, there would be at least a few attempts shown on the paper. :)

I think you may have lost a minus sign in your numerator.

My guess is this is your actual question:

(3 log(√2) + log 24 - log 3) / (log 512 - (1/2) log 64)

Try solving that instead.

Hint:

√2 = 2^½

512 = 2^9

64 = 2^6

a log(x) = log(x^a)

log(a) - log(b) = log(a/b)

log(a) + log(b) = log(ab)

If you need further help, gather six bits and I'll see what I can do.

UPDATE:

In case you missed it, "six bits" is equal to three quarters.

Answer:

3/4

- SlowfingerLv 62 months ago
(3 log(√2) + log 24 * log 3) / (log 512 - (1/2) log 64)

Notice there are several arguments of log here which are powers of 2, so let's try to factor out log(2)

log √2 = (1/2) log 2

log 24 = log (8 * 3) = log 8 + log 3 = 3log 2 + log 3

log 512 = log (2^9) = 9 log 2

log 64 = log (2^6) = 6 log 2

Now substitute all this

(3 log(√2) + log 24 * log 3) / (log 512 - (1/2) log 64) =

= (3 * (1/2) log 2 + log 3 (3log 2 + log 3)) / (9 log 2 - (6/2) log 2)

factor out log 2

= log 2 (3/2 + log 3 (3 + log 3 / log 2)) / (6 log 2)

log 2 cancels

= (3/2 + log 3 (3 + log 3 / log 2)) / 6

expand

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

I don't know what's the ultimate goal and where to stop but hope this helps.

- lenpol7Lv 72 months ago
Working to log base '10' for the calculator

(3log√2 + log24log3) / (log512-½log64)

Convert as much as possible to 'log(10)2'

Hence

(3/2)log2 + (log8 +log3)log3) / (9log2 - (1/2)6log2) =>

(3/2)log2 + ((log24)log3))/ (9log2 - 3log2)

((3/2) log2 + (log(24))log3) / ( 6log2)

NB I query the term log3 in the numerator. Should there be brackets in there somewhere.