Can Logarithmic Properties Simplify This Equation?

[diredragon]

Homework Statement

For clarification a have posted the equation below as a picture file.[/B]

Homework Equations

log(a*b)= loga + logb
log(a/b) = loga-logb
log(a^n) = nloga[/B]

The Attempt at a Solution

I don't know how to start. I can't remember the rule for powering the logarithms if there is any. Any hint would do...
Its got to do something with the fact that the numbers inside the logarithms go by some sort of a rule, 3 (+3), 6 (+6), 12 (+12), 24[/B]

 

[lordianed]

Hey diredragon, maybe you could try to post the picture file once more, I don't seem to be able to see it.

 

[diredragon]

My mistake sorry it seems it read error the last time. Can you see it now?

 

[lordianed]

Yes I can see it now, looks interesting, I'll have a go at it.

 

[Samy_A]

Homework Equations

log(a*b)= loga + logb
log(a/b) = loga/logb
log(a^n) = nloga[/B]

Second one is a typo, should be
log(a/b) = loga-logb

Hints:
$2^{log_2x}=x$

$log_26=log_23+log_22=log_23+1$
$log_212=log_23+log_24=log_23+2$
and so on ...
If you calculate the right side expression in terms of $log_23$, it simplifies nicely.

 

[diredragon]

Second one is a typo, should be
log(a/b) = loga-logb

Yup, missed that

 

[diredragon]

Yeah, i think i got it, the numbers do play a role as i can take out log2(3) for the other members as they all then can be if i apply the product rule like log2(6) = log2(2) + log2(3) = 1 + log(3). Ill have a go at it a bit later but ill appreciate it if you can try it so i can compare my results.

 

[Samy_A]

Ill have a go at it a bit later but ill appreciate it if you can try it so i can compare my results.

In case you want to compare with my result:

Spoiler

I got $2^A=72$ by simply calculating the right side expression using the hints given in my previous post. Maybe there is a smarter (and quicker) way to solve it.

 

[Nahcirn]

I'm curious, could you share exactly what you did when you say you got your answer by 'calculating the right side expression'? I feel like I've missed a short-cut and taken the long way around: I replaced log2(3) with x, so: 1/6( (x)^3 - (x + 1)^3 - (x + 2)^3 + (x + 3)^3 ), evaluated that, and ended up with 2x + 3.

So that left me with 2^[2(log2(3)) + 3]. I feel like that can be simplified, but I don't know how. I plug it into a calculator and I got the answer you got, so I know everything I've done so far is correct (but probably inefficient...)

 

[diredragon]

2^(2log2(3) + 3) = (2^3)*2(log2(3^2)) = 8*3^2 = 72
If it helps n*logb(c) = logb(c^n)

 

[Samy_A]

I'm curious, could you share exactly what you did when you say you got your answer by 'calculating the right side expression'? I feel like I've missed a short-cut and taken the long way around: I replaced log2(3) with x, so: 1/6( (x)^3 - (x + 1)^3 - (x + 2)^3 + (x + 3)^3 ), evaluated that, and ended up with 2x + 3.

That's exactly what I did (there may be a shortcut that escaped me).

So that left me with 2^[2(log2(3)) + 3]. I feel like that can be simplified, but I don't know how. I plug it into a calculator and I got the answer you got, so I know everything I've done so far is correct (but probably inefficient...)

You end up with $A=2*\log_23+3$
Then, using the properties of the logarithm given previously, and $2^{log_2y}=y$:
$2^A=2^{(2*\log_23+3)}=2^{\log_23^2}*2^3=3^2*2^3=9*8=72$

 

[diredragon]

It does seem like there is a shorter way out. Maybe there is a formula that calculates the sum of n consecutive cubes?

 

[Nahcirn]

2^(2log2(3) + 3) = (2^3)*2(log2(3^2)) = 8*3^2 = 72
If it helps n*logb(c) = logb(c^n)

For some reason I couldn't see that 2^log2(3^2) was the same thing as 3^2... It was easier to see when I turned the 3^2 into 9.

I just needed to remember that logarithms and exponents with the same base cancel each other out (which is the definition of a logarithm... Kind of important).

Thank you!

Still curious if there's a faster way, though.