Having problems with log proof.

[scguy]

I understand everything apart from the part that is circled in red, could somebody please explain how that is obtained from the information given.

Thanks.

[Dave]

It would help if you posted the original question

[Johnny5]

Let log_bx = m, and let log_ba = n.

 

Therefore:

 

x=b^m and a = bⁿ

 

So now, take the mth root of both sides of x=b^m to obtain:

 

x^1/m = (b^m)^1/m =b^m/m = b

 

So we now know that:

 

b = x^1/m and a = bⁿ, so by substitution it follows that:

 

a = (x^1/m)ⁿ

 

And multiplying the exponents it follows that:

 

a = x^n/m

 

Now, raising both sides to the power m, it follows that:

 

a^m = xⁿ

 

But go back one step to this result here:

 

a = x^n/m

 

If we raise both sides to the nth power we obtain:

 

aⁿ = x^nn/m

 

Therefore, the only way that:

 

aⁿ = x

is if

 

nn/m=1

 

From which it follows that m cannot be equal to zero. let it be the case that not (m=0). Now, multiply both sides by m to obtain:

 

nn=m

 

From which it follows that:

 

n²=m

 

From which it must follow that n is equal to the square root of m. That is:

 

n = m^1/2

 

So the line you circled in red is not unconditionally true, it is conditionally true.

 

If log_bx = m, and log_ba = n, and aⁿ = x

then

n = m^1/2

 

if log_bx = m, and log_ba = n, and not(n = m^1/2) then not(aⁿ = x).

 

Regards

[Dave]

Indeed. That's why I said the question might help (that condition might be in the question )

[scguy]

I thought it might be able to be worked from the information i provided but (i tried for a long time), i got the same answer u did. The whole proof seems to rely on the fact that n^2/m=1 which i am not really to happy about. I will post the rest of it later.

 

Thanks.

[scguy]

Ok, here is the result i am trying to prove:

 

log(base a)x= log(base B)x / log(base B)a.

 

Sorry about the notation but i didnt have time to write it properly.

[Dave]

Unless I'm mistaken, this is the change of base formula. Multiply through to get:

 

log_ax log_b a = log_b x.

 

Now suppose:

u = log_ax => a^u = x (1)

v = log_ba => b^v = a (2)

w = log_bx => b^w = x (3)

 

(1) and (3) together imply that: a^u = b^w.

(2) implies that: a^u = (b^v)^u = b^v.u

 

So we must have b^v.u = b^w => vu = w

 

Hence we are done.