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.

It would help if you posted the original question

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

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

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.

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.

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.