An Interesting Logarithmic System

[@RudeusGreyrat9]

My answer was 2ln2 ... Same thing, really

[@stvp68]

Cool graph!

[@trojanleo123]

ln4

[@SweetSorrow777]

I stick with the natural log as my catch-all logarithmic function except for base 10. That one gets common log. I find it strange WA uses common log as its' catch-all logarithmic function. w/e

[@Don-Ensley]

problem aˡⁿ ᵇ • b ˡⁿ ᵃ = 16 ln a • ln b = ? By properties of logs, take lns both sides and bring down exponents. Remember 16 = 2⁴. ln b ln a + ln a ln b = 4 ln 2 2 ln a • ln b = 4 ln 2 ln a • ln b = 2 ln 2 answer 2 ln 2

[@SidneiMV]

lnblna + lnalnb = 4ln2 *lnalnb = 2ln2 = ln4*

[@lesnyk255]

I never realized that a^ln(b) = b^ln(a)! Obvious in retrospect, but what isn't? I'll have to remember this. I did it a little differently. I let x = ln(a) & y = ln(b),, so a = e^x & b = e^y. This turns the original equation into [(e^x)^y][(e^y)^x] = 16. So (e^xy)^2 = 16, and e^xy = 4 (exponential disqualifies the negative root). Log both sides & back-substitute to get the final answer.

[@dlevi67]

Interesting that you hadn't realised the identity, when your first step gets (e^x)^y and (e^y)^x which are the same.

[@lesnyk255]

@@dlevi67"interestng", eh? Yes, I wasn't looking for it - had my eyes on the end goal

[@dlevi67]

@@lesnyk255 Totally understand (and I am often guilty of it too). I just found it funny - no offense meant! 😉

[@prollysine]

(a^lnb)^2=16 , a^lnb=+/-4 , b^lna=+/-4 , lna*lnb=ln4 ,