# Homework Help: How to solve this logarithmic equation?

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1. Nov 15, 2015

### diredragon

1. The problem statement, all variables and given/known data
For clarification a have posted the equation below as a picture file.

2. Relevant equations
log(a*b)= loga + logb
log(a/b) = loga-logb
log(a^n) = nloga

3. The attempt at a solution
I dont know how to start. I cant 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

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• ###### 20151115_150457.jpg
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Last edited: Nov 15, 2015
2. Nov 15, 2015

### lordianed

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

3. Nov 15, 2015

### diredragon

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

4. Nov 15, 2015

### lordianed

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

5. Nov 15, 2015

### Samy_A

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.

Last edited: Nov 15, 2015
6. Nov 15, 2015

### diredragon

Yup, missed that

7. Nov 15, 2015

### 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.

8. Nov 15, 2015

### Samy_A

In case you want to compare with my result:
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.

9. Nov 17, 2015

### 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...)

Last edited: Nov 17, 2015
10. Nov 17, 2015

### diredragon

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

11. Nov 18, 2015

### Samy_A

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

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

12. Nov 18, 2015

### diredragon

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

13. Nov 18, 2015

### Nahcirn

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.