+386 I have k,b and e > 0
log base b (k)= 2,5 and log base e (b) = 2
I have to find log base e (k) = ?
I tried this:
log base b (k)= log k/log b = 2,5/1
log base e (b) = log b/ log e = 2/1
If log base b (k)=2,5 then k=b^2,5
If log base e (b)=2 then b=e^2 Which means k= (e^2)^2,5=e^5
So now we have k= e^5 and b= e^2
log base e (k) = log base e (e^5) = log e^5 / log e = log e^4 = (approx) 1,73717
Am I right?
+386 Daaaamn never thought about the log base e (e)= 1 thing. So obvious.
Thanks a huge lot CPhill, you're helping me a lot.
+129840 logb k = 2.5 means that b^2.5 = k
loge b = 2 means that e^2 = b
So this means that b^2.5 = k → (e^2)^2.5 = k → e^5 = k
So
loge k = loge e^5 = 5 loge e = 5 * 1 = 5
+386 Daaaamn never thought about the log base e (e)= 1 thing. So obvious.
Thanks a huge lot CPhill, you're helping me a lot.
